This question requires an understanding of motion in two dimensions. The most important concept in this question is that the motion in each dimension is independent. Since the rock's initial velocity is purely in the horizontal direction, the initial velocity has no impact on the vertical velocity at any point. Likewise, the vertical acceleration has no impact on the horizontal speed. The rock will travel at  in the horizontal direction throughout its entire trajectory.

To solve this problem, the first step is to set up an expression for the horizontal and vertical velocities as a function of time:

Thus,

Substituting  for  and solving for  yields .

We first must find the initial vertical velocity () using:

We know that , since in 2 dimensions the vertical velocity at its max height is equal to zero.

we also know that  because that information is given, and that 

So plugging in what we know:

Knowing that  is constant and equals , we can use the pythagorean theorem to determine the Resultant initial Velocity:

First let's make a table of what we know:

 because it has zero velocity right when it is dropped.

It should also be noted it is simpler to define things so that the initial height is zero, so we don't have to deal with a bunch of negative numbers.

Also, since it is dropped vertically and we are only interested in the height it is dropped from we aren't interested in any information regarding the projectile's horizontal motion.

Let's use the equation:

since 

We can solve for , then plug in what we know:

This is our final answer.

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the equation

to solve for the time it takes the projectile to reach its max height.

Solving for :

Now lets plug in what we know and calculate the time:

, which is our final answer.

First let's write down the information we're given:

, since at a projectile's max height its vertical velocity equals zero.

Let's use the following equation to solve for the time it takes the projectile to reach its max height:

Solving for :

Now lets plug in what we know and calculate the time:

Since the projectile takes twice as long to land as its does to reach its max height we simply multiple the  we found by , which gives us  .

First let's write out the information we are given:

We know that in order to obtain the time it takes a projectile to hit the ground we just multiply  by , since the time it takes to reach the max height is half of the total time the project is in the air, so 

Now we can use the equation 

Since we are only concerned with the particle's motion in the horizontal direction, and we know that the horizontal velocity is constant . We also know  since that is the definition of the initial position.

This gives us:

Let's plug in what we have:

. This is the range and our final answer. The reason why we can just use  as the horizontal velocity is because the project is launched at a ° angle. 

There are many explanations for this. First you could simply insert a velocity at all of these angles and see which ends up with the greatest change in horizontal distance. You also could use basic calculus and solve for the greatest theta.

This is a projectile motion problem. The problem states that it was thrown off the building at an angle of 30 degrees to the horizontal. So before we can find time, we need to find the horizontal and vertical parts of this velocity. 

Since   and 

It follows that 

Now, before we can find far it went, we need to find how long it was in the air. We need to solve for time.

With the equation   we can find t.

X is the initial height which in this case is the  tall building. V is the initial vertical velocity  and  

Plugging all that in and solving yields 

Knowing that the time the ball is in the air is  and the constant horizontal velocity is , we can plug in these known values into the simple distance formula to solve for distance.

The angular frequency of a simple pendulum is , where is the length of the pendulum.

For this question, we need to recall what simple harmonic motion is. Remember that it is a periodic motion where the restoring force depends on the displacement of the object undergoing these motions. So to answer this question, we need to keep this idea in mind and see which example doesn't match up.

A mass on a pendulum moving back and forth is clearly an example of simple harmonic motion. As the mass moves further from the center in either direction, it experiences a greater and greater force in the opposite direction.

A child swinging on a swing set is another correct example. This situation is analogous to the mass on a pendulum swinging back and forth.

A vibrating guitar string is yet another example of simple harmonic motion. After it is plucked, the string oscillates back and forth.

Finally, a book falling to the ground does not represent harmonic motion. Once the book is released from rest, we intuitively know that it will fall to the ground and will then stay there; in no way is there any periodic motion.