We know that the plane needs to drop 3,000 meters in 10 minutes. Therefore, we can calculate the vertical velocity that the plane needs to be traveling at:

Knowing the plane's velocity, we can caluclate what angle at which it needs to decend to achieve this vertical velocity:

Rearranging for theta:

Plugging in our values:

We can use our kinematics equations to determine the height of the rocket the moment it loses thrust:

If we clarify the initial state being the moment the rocket loses thrust and the final state being the point of its maximum height, we can use the expression for conservation of energy to calculate the final height:

Removing final kinetic energy (which will be zero) and substituting in our expressions for each variable:

Cancel out mass and rearrange for final height:

From the problem statement, we can determine the initial velocity:

Now that we have values for all of our variables, we can solve:

There are two ways to solve this problem. The first and much easier way is to use the range equation. The second is using your kinematics equations.

Method 1: Range Equation

The range equation is the following:

We know everything ecvept for the initial velocity. However, we can calculate it knowing that the cannon transfers 500 J of energy to the ball. Therefore:

Rearranging for velocity:

Now we can plug everything into the range equation:

Method 2: Kinematics Equations

As in the first method, we can calculate the initial velocity of the ball:

Rearranging for velocity:

We can then split this into it's components:

We can use the y-component to calculate how long the ball is in the air. We can do one of two things:

1. Calculate how long the ball takes to reach it's peak and then multiply by 2
2. Calculate how lon gthe ball takes to get a velocity of -5

We'll go with method 2:

Rearranging for t:

Then we can multiply this time by the horizontal velocity (which stays constant because we are neglecting air resistance).

To find the acceleration of the box traveling down the incline, the mass is not needed. Using the incline of the plane as the x-direction, we can see that there is no movement in the y-direction; therefore, we can use Newton's second, F = ma, in the x-direction.

There is only one force in the x-direction (gravity), however gravity is not just equal to “mg” in this case. Since the box is on an incline, the gravitational force will be equal to mgsin(30^o). Substituting force into F  =ma we find that mgsin(30^o) = ma. We can now cancel out masses and solve for acceleration.

There are two forces in play in this scenario: friction and gravity. We can use Newton's second law to develop an expression for the acceleration of the block:

Substituting in expressions for each force, we get:

Canceling out mass and rearranging for acceleration, we get:

We know all of our values, allowing us to solve for the acceleration:

We can use the equation for conservation of energy to solve this problem:

If we assume the initial state is when the slingshot is fully loaded and the final state is at the ball's maxmium height, we can completely eliminate kinetic energy to get:

Substituting in our value for initital potential energy and an expression for final potential energy, we get:

Rearranging for final height we get:

The basic kinematics equation for motion under constant acceleration is:

We do not have the acceleration, but we can find it:

Apply this to the first equation and solve.

The equation for speed is:

The runner has gone around the track 1.5 times, thus his total distance traveled is:

We can find the elapsed time from the walker: Rearrange the speed equation to find time:

Now we have all the information we need to find the runner's speed:

If an object is moving to the right, then its displacement is positive. If its displacement is positive then its velocity must be positive as well. The object is speeding up. This implies that the magnitude of its velocity is increasing. If this is true we have that the velocity of the object, , after some time, , has passed must be a larger positive velocity than say, , the object's velocity before that time passed. Therefore, the object's average acceleration is given by:

 since 

So we have that the object's acceleration is positive. If the object's acceleration is positive, then there must be a positive net force causing the acceleration, due to Newton's first law: . Therefore, it is false that the object experiences a net force of 0N.

By definition, acceleration is the rate of change of velocity with respect to time. The average acceleration of an object is given by:

Therefore, if velocity is constant it must be the case that acceleration is 0, since the initial and final velocities of the object are equal. Similarly, if acceleration is non-zero, then it must be the case that the initial and final velocities of the object are different (so velocity is changing). Now, when an object's acceleration and initial-velocity have the same sign (i.e. the same direction) then the object will speed up. In other words, the absolute value of its velocity (the magnitude) increases. This is easy to see if we solve for the final velocity in our equation:

If both  and  have the same sign, then the absolute value of  will be greater as time passes, meaning the object is speeding up. (Try plugging in some numbers). Similarly, when  and  have opposite signs, the magnitude of the object's velocity must start to decrease. After enough time it might even come to a stop, and if it continues to experience the acceleration it will change direction and start gaining velocity on the opposite direction. Regardless, it must first slow down (think about when you throw an object on the air).

Example: 

Example: 

Remember that the positive and negative signs indicate the direction of the vectors. So when it comes to slowing down or speeding up, we want to look at the magnitude of the velocity. Therefore, the only statement that is false is the one that says that if an object has both negative initial velocity and acceleration it will slow down. The object will speed up and continue to move to the left.