Let's first write down the information we are given:

In order to solve this problem we must apply the conservation of energy, which states  since no friction.

This means that as the project reaches its max height energy is converted from Kinetic energy (energy of motion) to potential gravitational energy (based off of height).

We can subtract  from  to get the    at its max height

=

so we can solve for the mass

where 

therefore 

The formula for gravitational potential energy is:

To find the potential energy lost, we need to find the potential energy of the ball at two heights, then find the difference. 

Note that the energy was not actually lost; rather, it was converted to kinetic energy. 

Gravitational potential energy is given by the equation:

We are given all the information needed to solve for the potential energy.

Plug in known values and solve.

Recall that the units for energy are Joules. Newtons is the unit for force.

The equation for gravitational potential energy is:

We are given all the information needed to answer the question.

Plug in known values and solve.

For this question, we're told than an object is travelling with a certain velocity towards a spring. After colliding with the spring and causing it to undergo a maximum displacement, we're asked to provide an expression for the spring constant.

In order to answer this question, we'll need to consider the energy of the object and the spring during the process described. First, the object is moving with a certain velocity towards the spring. Thus, the object has kinetic energy. When the object collides with the spring and displaces the spring to its maximum displacement, there will be a brief instant where the object has completely stopped moving. At this point, all of the block's kinetic energy will have been transferred into the spring in the form of potential energy. With this information in mind, we can equate the two forms of energy.

Now, we can rearrange terms in order to isolate .

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  .