Motion in Two Dimensions - AP Physics 1
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A home run derby is being held in a stadium where the home run wall is 
from home plate. The wall itself has a height of 
. The stands begin at the wall and with equal height, and rise at an angle of 
to the horizontal.
Neglect air resistance and assume 
A batter hits a ball from a height of 
with an initial velocity of 
and an angle of 
. What is the horizontal distance traveled before it colliding with the stands/field?
A home run derby is being held in a stadium where the home run wall is
Neglect air resistance and assume
A batter hits a ball from a height of
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We can use the following piecewise function to determine the height of the ball as a function of the horizontal distance it has traveled:
If 
, then 
If 
, then 
First, we can guesstimate if the ball will land in the field in the stands or the field using the range equation:
Plugging in our values:
This means that the ball will reach its original height about 58 meters past the home run wall. Therefore, let's operate on the idea that the ball lands in the stands.
We can begin with the following kinematic equation:
Where we can define time as :
Where x is the horizontal distance traveled before the ball lands.
Substituting in expressions for our component velocities, we get:
Then substituting the 2nd expression into the first:
Now we can use the piecewise function above to replace final height or final distance. It will be much simpler to replace final height, so we'll do that:
Expanding the expression a bit:
Then rearranging, we get:
Plugging in values, we get:
Using the quadratic equation, we get:
Obviously our distance will be positive, so we will go with the second one:
We can use the following piecewise function to determine the height of the ball as a function of the horizontal distance it has traveled:
If
If
First, we can guesstimate if the ball will land in the field in the stands or the field using the range equation:
Plugging in our values:
This means that the ball will reach its original height about 58 meters past the home run wall. Therefore, let's operate on the idea that the ball lands in the stands.
We can begin with the following kinematic equation:
Where we can define time as :
Where x is the horizontal distance traveled before the ball lands.
Substituting in expressions for our component velocities, we get:
Then substituting the 2nd expression into the first:
Now we can use the piecewise function above to replace final height or final distance. It will be much simpler to replace final height, so we'll do that:
Expanding the expression a bit:
Then rearranging, we get:
Plugging in values, we get:
Using the quadratic equation, we get:
Obviously our distance will be positive, so we will go with the second one:
An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?
An object is shot from the ground at 75m/s at an angle of 45⁰ above the horizontal. How high does the object get before beginning its descent?
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The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use 
.
Next we find how long it takes to reach the top of its trajectory using 
.
t = 5.3s
Finally, find how high the object goes with 
.
The velocity must be broken down into x (horizontal) and y (vertical) components. We can use the y component to find how high the object gets. To find vertical velocity, vy, use
Next we find how long it takes to reach the top of its trajectory using
t = 5.3s
Finally, find how high the object goes with
An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?
An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?
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First, find the horizontal (x) and vertical (y) components of the velocity
Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.
t = 6.25s
Total time in the air is therefore 12.5s (twice this value).
Finally, find distance traveled my multiplying horizontal velocity and time.
First, find the horizontal (x) and vertical (y) components of the velocity
Next, find how long the object is in the air by calculating the time it takes it to reach the top of its path, and doubling that number.
t = 6.25s
Total time in the air is therefore 12.5s (twice this value).
Finally, find distance traveled my multiplying horizontal velocity and time.
A 2kg box is at the top of a frictionless ramp at an angle of 
. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
When the box is released, how long will it take the box to reach the ground?
A 2kg box is at the top of a frictionless ramp at an angle of
When the box is released, how long will it take the box to reach the ground?
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We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation 
.
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of 
. Since the angle is 60o, the acceleration on the box is 
Finally, we can plug these values into the following distance equation and solve for time.
We can start this problem by determining how far the box will travel on the ramp before hitting the ground. Since the vertical distance is 30m and we know the angle of the ramp, we can determine the length of the hypotenuse using the equation
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity. Because the box is on a sloped surface, the box will not experience the full acceleration of gravity, but will instead be accelerated at a value of
Finally, we can plug these values into the following distance equation and solve for time.
A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
What is the velocity of the box just before it hits the ground?
A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.
What is the velocity of the box just before it hits the ground?
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We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation 
.
We can plug these values into the following distance equation and solve for time.
Now that we know the acceleration on the box and the time of travel, we can use the equation 
to solve for the velocity.
We can start this problem by determining how much time it takes the box the reach the ground. Since the vertical distance is 30m and we know the angle of the ramp.
Now that we have the distance traveled, we can determine the acceleration on the box due to gravity, using the equation
We can plug these values into the following distance equation and solve for time.
Now that we know the acceleration on the box and the time of travel, we can use the equation
A ball gets pushed off a 
high table with a horizontal speed of 
. How far does the ball travel horizontally before hitting the ground?
A ball gets pushed off a
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This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.
Choosing the ground to be the zero height, we have 
and 
.
Also, knowing that the initial vertical velocity is zero, we know that 
.
The kinematic equation simplifies using these values.
Rearrange the equation to isolate time.
We know that 
is the acceleration due to gravity: 
. Plug in the values to solve for time.
We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation 
.
We know that 
and that 
. Using these values and the time, we can solve for the horizontal distance travelled.
This is a two step problem. The first step is to calculate the time it takes for the ball to reach the ground. To find this time, we use the following kinematic equation dealing with vertical motion.
Choosing the ground to be the zero height, we have
Also, knowing that the initial vertical velocity is zero, we know that
The kinematic equation simplifies using these values.
Rearrange the equation to isolate time.
We know that
We now have the time the ball is travelling before it hits the ground. Use this value to find the horizontal distance before it hits the ground with the kinematic equation
We know that
A car drives north at 
for 
, then turns east and drives at 
for 
. What is the magnitude and direction of the average velocity for the trip?
A car drives north at
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First, determine how far the car travels in each direction:
Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.
Find the average velocity by dividing the total displacement by the total time.
Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.
Use the north and the east directional displacements to find the angle.
Our final answer will be:
First, determine how far the car travels in each direction:
Now that we have the directional displacements, we can find the total displacement by using the Pythagorean Theorem.
Find the average velocity by dividing the total displacement by the total time.
Velocity is a vector, meaning it has both magnitude and direction. Now that we have the magnitude, we can find the direction by using trigonometry.
Use the north and the east directional displacements to find the angle.
Our final answer will be:
A ball is launched at an angle of 
above the horizontal with an initial velocity of 
. At what time is its vertical velocity 
?
A ball is launched at an angle of
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Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.
We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.
Using this in our kinematics formula, we solve for the time.
Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.
Any projectile has a vertical velocity of zero at the peak of its flight. To solve this question, we need to find the time that it takes the ball to reach this height. The easiest way is to solve for the initial vertical velocity using trigonometry, and then use the appropriate kinematics equation to determine the time.
We know that the final vertical velocity will be zero. We can solve for the initial vertical velocity using the given angle and total velocity.
Using this in our kinematics formula, we solve for the time.
Keep in mind that this is only the vertical velocity. The total velocity at the peak is not zero, since the ball will still have horizontal velocity.
A plane is traveling from Portland to Seattle, which is 100 miles due north of Portland. There is a constant wind traveling southeast at 30 mph. If the plane needs to get to Seattle in one hour while flying due north, at what speed (relative to the wind) and angle should the pilot fly?
A plane is traveling from Portland to Seattle, which is 100 miles due north of Portland. There is a constant wind traveling southeast at 30 mph. If the plane needs to get to Seattle in one hour while flying due north, at what speed (relative to the wind) and angle should the pilot fly?
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First we need to find out at what the speed relative to the ground the plane needs to fly. The plane needs to cover 100 miles in 1 hour, so it's simply 100 mph due north.
Now we need to calculate its speed relative to the wind and its angle. We know the sum of the wind vector and plane vector needs to equal 100 mph due north. Therefore, all east/west movement must cancel out and all north/south movement must add to 100mph.
We can separate the wind velocity into its components:
We can also represent the components of the plane's velocity:
If the trig functions seem reversed, this is because the angle in question is between the y-coordinate and a vector pointing slightly west of north.
We can also represent the components of the velocity relative to the ground:
Since the horizontal velocity is equal to 0, we can set the x-components of the wind and plane vectors equal to each other:
The sum of the y-components of the wind and plane vectors must equal 100:
The wind vector is subtracted because it is in the opposite direction of the plane vector.
Now we just need to isolate a variable and substitute one equation into another. We will isolate the total plane velocity in the first equation:
Substituting this into the second equation, we get:
Solving for theta, we get 
(west of north)
We can plug this into the first equation to get:
First we need to find out at what the speed relative to the ground the plane needs to fly. The plane needs to cover 100 miles in 1 hour, so it's simply 100 mph due north.
Now we need to calculate its speed relative to the wind and its angle. We know the sum of the wind vector and plane vector needs to equal 100 mph due north. Therefore, all east/west movement must cancel out and all north/south movement must add to 100mph.
We can separate the wind velocity into its components:
We can also represent the components of the plane's velocity:
If the trig functions seem reversed, this is because the angle in question is between the y-coordinate and a vector pointing slightly west of north.
We can also represent the components of the velocity relative to the ground:
Since the horizontal velocity is equal to 0, we can set the x-components of the wind and plane vectors equal to each other:
The sum of the y-components of the wind and plane vectors must equal 100:
The wind vector is subtracted because it is in the opposite direction of the plane vector.
Now we just need to isolate a variable and substitute one equation into another. We will isolate the total plane velocity in the first equation:
Substituting this into the second equation, we get:
Solving for theta, we get
We can plug this into the first equation to get:
A baseball is traveling with a velocity of 
at an angle of 
above horizontal. What is the velocity of the ball after two seconds?
A baseball is traveling with a velocity of
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To solve this problem, we first need to split the velocity into its vector components.
Initial vertical velocity:
Initial horizontal velocity:
Since we are neglecting air resistance, horizontal velocity does not change over time. We only need to calculate the new vertical velocity after two seconds, using kinematics:
The negative sign simply means that the vertical velocity has changed direction, and is now pointed downward.
We can use the following equation to determine the total velocity, which will be the sum of the horizontal and vertical velocity vectors:
To solve this problem, we first need to split the velocity into its vector components.
Initial vertical velocity:
Initial horizontal velocity:
Since we are neglecting air resistance, horizontal velocity does not change over time. We only need to calculate the new vertical velocity after two seconds, using kinematics:
The negative sign simply means that the vertical velocity has changed direction, and is now pointed downward.
We can use the following equation to determine the total velocity, which will be the sum of the horizontal and vertical velocity vectors:
Two balls are thrown off the edge of a cliff of height 
from the same location. Both balls are thrown with an initial velocity of 
. The first ball is thrown horizontally and the second ball is thrown at and angle of 
above the horizontal. When the cliff is a particular height, let's call it 
, the two balls will land at the same spot on the ground.
Which of the following is true?
Two balls are thrown off the edge of a cliff of height
Which of the following is true?
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If 
, the horizontally thrown ball will travel no distance. As the height of the cliff increases, the horizontally thrown ball will have more and more air time until it begins to travel further than the ball thrown 
above the horizontal. This is because the distance each ball will go equals the time it's in the air multiplied by its horizontal velocity (since it's not accelerating in the x-direction). The time the ball thrown 
above the horizontal is always the same amount greater than the time the ball thrown horizontally is in the air. The ball thrown horizontally has a horizontal velocity greater than the ball thrown 
above the horizontal (by a factor of 
). This means the ball thrown horizontally travels a distance of 
while the ball thrown 
above the horizontal travels a distance of 
. If we equate these two expressions, we get 
. When 
is less than this, the ball thrown 
above the horizontal travels further and when 
is greater the ball thrown horizontally travels further. 
represents the time is takes for the ball thrown horizontally to hit the ground when the height of this cliff is 
.
If
A cannon ball is fired at an angle of 45 degrees from the horizontal. Which of the following describes the kind of motion experienced by the cannon ball in the y-axis and the x-axis respectively?
A cannon ball is fired at an angle of 45 degrees from the horizontal. Which of the following describes the kind of motion experienced by the cannon ball in the y-axis and the x-axis respectively?
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The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.
The cannon ball has a two-dimensional motion, specifically projectile motion. Any two-dimensional motion consists of two one-dimensional motions happening at the same time. This is why we break vectors into components, so we can analyze each one-dimensional motion separately. The motion in the y-axis and the x-axis are connected by the time variable, since they happen at the same time. For projectile motion, we have that on the y-axis (the vertical axis) a constant acceleration (gravity) is acting on the cannon ball. For the y-axis motion with constant acceleration or free fall (which is motion with constant acceleration where the acceleration is gravity) are correct choices. On the x-axis (the horizontal axis), there is no acceleration acting on the cannon ball horizontally. Therefore, on the x-axis the cannon ball experiences motion with constant velocity. Then the only answer choice that describes the motion of the cannon ball in each axis correctly is: y-axis: motion with constant acceleration and x-axis: motion with constant velocity.
Let 
represent the magnitude of the gravity acceleration.
As in the given figure, a ball is kicked from the ground giving it an initial velocity of 
at an angle of 
. It takes a time 
to reach the highest point in its trajectory. When it finally reaches the ground, which is perfectly flat, the ball covers a range (the total displacement on the horizontal axis) of 
. In terms of 
, 
, 
, and 
, how much time 
does it take the ball to reach its highest point?
Let
As in the given figure, a ball is kicked from the ground giving it an initial velocity of
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How much time it takes the ball to reach its highest point depends on its motion in the y-axis. Therefore, it depends on the ball's initial velocity in the y-direction and the acceleration of gravity. Note that at the highest point of its trajectory, the ball reaches a velocity of 
in the y-direction for an instant, just before it changes direction and starts coming down. Therefore we can use the following kinematic equation for motion with constant acceleration to find 
:
Here, 
is the final velocity on y for this part of the ball's trajectory (i.e. its velocity at the highest point), 
is the ball's initial velocity in the y-direction, and 
is the acceleration acting on the y-axis (gravity). We can express 
in terms of 
and 
by obtaining the y component of the initial velocity of the ball, which is given by:
Note that the acceleration in y-direction is negative since gravity pulls downwards.
Solve for 
.
How much time it takes the ball to reach its highest point depends on its motion in the y-axis. Therefore, it depends on the ball's initial velocity in the y-direction and the acceleration of gravity. Note that at the highest point of its trajectory, the ball reaches a velocity of
Here,
Note that the acceleration in y-direction is negative since gravity pulls downwards.
Solve for
Let 
represent the magnitude of gravity acceleration.
As in the given, figure, a ball is kicked from the ground giving it an initial velocity of 
at an angle of 
. It takes a time 
to reach the highest point in its trajectory. When it finally reaches the ground, which is perfectly flat, the ball covers a range (the total displacement on the horizontal axis) of 
. Which of the following equations represents the range, 
, in terms of 
, 
, 
, and 
?
Let
As in the given, figure, a ball is kicked from the ground giving it an initial velocity of
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An important thing to notice is that this projectile motion is symmetrical, meaning that the ball starts and ends its motion at the same height. Then, it must be the case that the ball takes the same amount to reach its highest point from its original position as it takes the ball to fall down from the highest point to its final position. In other words, it takes a time of 
to complete the entire trajectory. Since the ball's movement follows projectile motion, it experiences motion with constant velocity on the horizontal axis (the x-axis). We can calculate this velocity by obtaining the x-component of the ball's initial velocity:
Since the velocity is constant, we can express it in terms of displacement in the x-direction and time with the following equation:
Here, 
is the displacement in the x-direction and 
is the time.
Therefore, we have that for the range (the ball's entire displacement in the x-direction for its motion) is given by:
An important thing to notice is that this projectile motion is symmetrical, meaning that the ball starts and ends its motion at the same height. Then, it must be the case that the ball takes the same amount to reach its highest point from its original position as it takes the ball to fall down from the highest point to its final position. In other words, it takes a time of
Since the velocity is constant, we can express it in terms of displacement in the x-direction and time with the following equation:
Here,
Therefore, we have that for the range (the ball's entire displacement in the x-direction for its motion) is given by:
Suppose that a golf ball is struck such that it travels at a speed of 
at an angle 
to the horizontal. Neglecting air resistance, how long will the golf ball remain in the air before it touches the ground again?
Suppose that a golf ball is struck such that it travels at a speed of
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We're told that the golf ball is starting its parabolic journey with a certain velocity at an angle to the ground. To solve for the time the golf ball stays airborne, we'll need to consider the x and y-components of the ball's trajectory.
First, we'll need an expression that considers the ball's velocity in the x-direction.
We'll also need an expression for the y-component of the velocity.
We'll need an expression that can relate the vertical distance traveled with the time spent in the air. Since the only acceleration occurring in this scenario is due to gravity, the result is that acceleration is constant. Therefore, we can make use of some of the kinematics equations:
It's also important to note that once the ball lands back on the ground, only its horizontal displacement will have changed, while its vertical displacement will remain unchanged.
We also have to remember that in this case, the source of acceleration is from gravity, which points downwards. If we define up as the positive y direction, then down must be the negative y direction. Therefore, we can write:
Plug in the vertical component of velocity and solve for time.
We're told that the golf ball is starting its parabolic journey with a certain velocity at an angle to the ground. To solve for the time the golf ball stays airborne, we'll need to consider the x and y-components of the ball's trajectory.
First, we'll need an expression that considers the ball's velocity in the x-direction.
We'll also need an expression for the y-component of the velocity.
We'll need an expression that can relate the vertical distance traveled with the time spent in the air. Since the only acceleration occurring in this scenario is due to gravity, the result is that acceleration is constant. Therefore, we can make use of some of the kinematics equations:
It's also important to note that once the ball lands back on the ground, only its horizontal displacement will have changed, while its vertical displacement will remain unchanged.
We also have to remember that in this case, the source of acceleration is from gravity, which points downwards. If we define up as the positive y direction, then down must be the negative y direction. Therefore, we can write:
Plug in the vertical component of velocity and solve for time.
An object with mass 
slides down an incline plane with slope 
and height 
. What does the coefficent of friction 
have to be in order for the object to not gain velocity?
An object with mass
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In order for the object not to gain velocity, all the forces acting on it need to cancel out. The normal force 
will already cancel with the perpendicular force due to gravity (
) by definition. In order to find the correct coefficient of friction, we set the magnitude of the force due to friction 
equal to that of the parallel force due to gravity 
:
The mass and gravity cancel here.
In order for the object not to gain velocity, all the forces acting on it need to cancel out. The normal force
The mass and gravity cancel here.
An object weighing 15kg is fired from 30m above the ground at an angle of 
to the horizontal with an initial velocity of 
. Determine how long it takes the object to reach its maximum height. Ignore friction.
An object weighing 15kg is fired from 30m above the ground at an angle of
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At the projectile's maximum height, we know that it's vertical velocity will be 
Since the only type of acceleration the object faces is the acceleration due to gravity, we can model its velocity using only two terms: it's initial velocity and acceleration due to gravity. Recall that the formula for velocity with constant acceleration is given as:
Here, 
is the object's constant acceleration, 
is the initial velocity, and 
is time.
The only acceleration is acceleration due to gravity, which is:
To determine initial velocity 
, we have to determine a few other things. The object is being launched at a speed of 
at an angle of 
. Therefore, the vertical component (the vertical velocity of this object) is going to be
Therefore, we can model the velocity of this object as:
The signs of the two terms must be opposite because they are going in different directions. Negative velocities in this model mean the object is moving towards the center of the earth.
Since the object reaches its maximum point when 
we must solve the following equation for 
At the projectile's maximum height, we know that it's vertical velocity will be
Here,
The only acceleration is acceleration due to gravity, which is:
To determine initial velocity
Therefore, we can model the velocity of this object as:
The signs of the two terms must be opposite because they are going in different directions. Negative velocities in this model mean the object is moving towards the center of the earth.
Since the object reaches its maximum point when
A 2 kg cat is stuck in a 10 meter high tree. Just as a fire fighter atttempts to save the cat, the cat slips and falls out of the tree. Luckily, Superman is standing only 3 kilometers away and sees the cat fall. Assuming Super man has instananeous reactions and begins running towards the cat just as it falls, how fast does Superman have to run in order to catch the cat before it hits the ground?
A 2 kg cat is stuck in a 10 meter high tree. Just as a fire fighter atttempts to save the cat, the cat slips and falls out of the tree. Luckily, Superman is standing only 3 kilometers away and sees the cat fall. Assuming Super man has instananeous reactions and begins running towards the cat just as it falls, how fast does Superman have to run in order to catch the cat before it hits the ground?
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Superman needs to cover 3,000 meters. But the question is how much time does he have to cover those 3,000 meters? If you know the time, you can just divide 3,000 meters by the time he is traveling and you have his speed! But how do we know how quickly he needs to cover the 3,000 meters? The falling cat tells us! The acceleration due to gravity is a constant on the surface of Earth and it will dictate how quickly something hits the ground once dropped. Considering we ignore air resistance in physics 1, we don't care about the object's size or mass. So all we need to do is find the appropriate kinematic equation(these equations relate displacement, velocity, acceleration and time) and solve for time. The way I chose a kinematic equation is I organize variables into things we are told in the problem and things we are not told in the problem. In this case, we are given information about the height of the cat, it's mass and that it slipped (velocity in both x and y directions is zero). Not much to go off of but it is all we need! We are solving for time and we have information about initial velocity and displacement.
The equation:
We can see with this equation that under the assumption that the initial velocity is zero, it simplifies to a relationship between x and t (acceleration is g and is a constant). Rearranging for t gives:
Plugging in the appropriate numbers, the time that it take for the cat to hit the ground will be 1.43 seconds. Now we know Superman must cover the 3,000 in 1.43 seconds.
Dividing 3,000 by 1.43 yelids a velocity of
.
Superman needs to cover 3,000 meters. But the question is how much time does he have to cover those 3,000 meters? If you know the time, you can just divide 3,000 meters by the time he is traveling and you have his speed! But how do we know how quickly he needs to cover the 3,000 meters? The falling cat tells us! The acceleration due to gravity is a constant on the surface of Earth and it will dictate how quickly something hits the ground once dropped. Considering we ignore air resistance in physics 1, we don't care about the object's size or mass. So all we need to do is find the appropriate kinematic equation(these equations relate displacement, velocity, acceleration and time) and solve for time. The way I chose a kinematic equation is I organize variables into things we are told in the problem and things we are not told in the problem. In this case, we are given information about the height of the cat, it's mass and that it slipped (velocity in both x and y directions is zero). Not much to go off of but it is all we need! We are solving for time and we have information about initial velocity and displacement.
The equation:
We can see with this equation that under the assumption that the initial velocity is zero, it simplifies to a relationship between x and t (acceleration is g and is a constant). Rearranging for t gives:
Plugging in the appropriate numbers, the time that it take for the cat to hit the ground will be 1.43 seconds. Now we know Superman must cover the 3,000 in 1.43 seconds.
Dividing 3,000 by 1.43 yelids a velocity of
A cannon is fired from the top of a 
cliff as shown in the figure below. If the cannonball travels a horizontal distance 
of 
in 
, at what angle was the cannon fired? (Hint: Find an expression for 
. You do not need the initial velocity to solve this.)
A cannon is fired from the top of a
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At first this problem could seem challenging, but in fact, it is straight-forward if you can remove the initial velocity from the equations. We begin by writing the equations of motion for the x and y directions.
We then devide the two equations to get an expression for the tangent of the angle, as well ass remove the initial velocity from the problem. This enables us to use the information given to calculate the angle the projectile was fired.
At first this problem could seem challenging, but in fact, it is straight-forward if you can remove the initial velocity from the equations. We begin by writing the equations of motion for the x and y directions.
We then devide the two equations to get an expression for the tangent of the angle, as well ass remove the initial velocity from the problem. This enables us to use the information given to calculate the angle the projectile was fired.
A projectile from a hand gun is fired at 
above the horizontal at a speed of 
. How many seconds is the projectile airborne? Assume 
and that the hand gun was fired a negligible distance from the ground.
A projectile from a hand gun is fired at
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Since the question is asking how long the projectile was airborne, we are dealing with motion in y-direction only. Find initial velocity in the y-direction and substitute into the kinematic equation:
Since the question is asking how long the projectile was airborne, we are dealing with motion in y-direction only. Find initial velocity in the y-direction and substitute into the kinematic equation: