To solve this problem, we'll need to consider the initial energy of the block. Since it is starting from rest, it will not have any kinetic energy. Rather, all of its energy will be in the form of gravitational potential energy.

Once the block is released from its resting position, it will begin to fall towards the ground. As it falls, it will lose potential energy, but it will gain kinetic energy as it picks up speed. What's important to realize is that because we are neglecting air resistance, we have a situation in which total mechanical energy is conserved. Therefore, as the block loses potential energy during its fall, it will gain an exactly equal amount of kinetic energy. Therefore, no matter where the block is during its fall, its total mechanical energy will be the same! Keeping this in mind, we can equate the total mechanical energy of the block at any point during its fall with its initial potential energy, which we calculated above.

To begin with, we'll need to approach this problem from two angles. First, we'll need to consider the changes in energy that occur to the block as it slides from its initial position at rest to the point at which it falls off. Since we are told that the semicircle is frictionless, then we can expect the energy in the block-semicircle system to remain constant - the total mechanical energy will not change. Additionally, since the block is moving in a circular motion, we'll need to consider the centripetal force acting upon the block.

So, we need to find an expression that relates the initial energy to the final energy. In doing so, we'll need to consider both the gravitational potential energy of the block, as well as its kinetic energy. Initially, the block has potential energy but it does not have kinetic energy because it is at rest. Upon falling off the semicircle, the block will have both potential and kinetic energy.

Because the block is initially at the top of the semicircle, this starting height is equal to the radius of the semicircle. Once it slides down, it will be at a height that is less than its initial height. We can relate the height at any given point along the semicircle to its radius by employing trigonometry as shown below:

Plug in these expressions for height into the above expressions for energy.

Solve for velocity.

But we also need to approach this problem from another perspective. Since the block is moving in a circular motion, we know that we will be dealing with a centripetal force. Thus, it will be useful for us to consider the forces acting upon the block as it slides down the semicircle.

First, let's consider the forces acting parallel to the surface of the semicircle. Since the surface is frictionless, we won't have to worry about any frictional forces as the block slides. That means that our only force in this direction is the gravitational component that acts tangentially to the surface. Fortunately, we will not need this value to find our answer.

Now, let's consider the forces acting perpendicularly to the surface, or in other words, radially. For one, we have the radial component of the gravitational force, which points towards the center of the circle. Acting in the opposite direction to this we have the normal force. The difference in these two values is what provides the centripetal force that keeps the block on the surface.

In order to identify the height at which the block loses contact with the surface, we need to realize that this will happen once the radial component of the gravitational force can no longer provide an adequate centripetal force to keep the block attached. In such a situation, the block will lose contact with the surface and the normal force will reduce to zero. We will then have:

Now, if we substitute the expression for velocity that we derived from considering energy into the expression for centripetal force, we obtain:

By canceling common terms and rearranging, we obtain the following expression:

To begin with, we'll need to consider the energy of the blocks initially and right before the collision. Initially, block 1 is starting at some height and therefore has gravitational potential energy. Block 2 is initially not moving and is at the bottom of the half pipe, therefore it does not have any energy.

We'll also need to consider the energy of the blocks the instant right before the collision. In such a situation, all of the gravitational potential energy of block 1 has been converted into kinetic energy. Block 2, just as before, is at rest at the bottom of the half pipe and thus does not have any energy.

From these equations and knowing that energy is conserved, we can solve for the velocity of block 1 just before the collision.

In addition to considering the energy associated with the collision, it's also important for us to consider the momentum before and after the collision. In the instant right before the collision, block 1 has a velocity which we calculated above, while block 2 is at rest. After the collision, we're told in the question stem that the two blocks will move together. And due to conservation of momentum, we can relate the initial and final momentum as equal to each other.

Now solve for the final velocity of the two-block system.

Plug in the value for the velocity of block 1.

The above expression represents the velocity of the two-block system right after the collision. Knowing this, we can find an expression for the energy of both blocks the instant after the collision, and then relate this value to the final energy of both blocks as they ascend.

The instant after the collision, both blocks will be at the bottom of the halfpipe and will thus have no gravitational potential energy. However, the two blocks will have a certain velocity, which we calculated above. Thus, the two blocks will have kinetic energy. As the two blocks ascend and reach their maximum height, all of their kinetic energy will be converted into gravitational potential energy.

Now, plug in the final velocity expression that we calculated above for the two blocks.

While this problem can be solved by using kinematic equations for motion with constant acceleration, the simplest solution comes from using conservation of energy. Immediately after John leaves the ground, he has a potential energy of 0J and a kinetic energy given by his initial velocity. So we have that John's initial energy is given by:

When John reaches the highest point, his velocity must be zero since this would be the point where he stops for an instant just before start falling down. Therefore, at his highest point John has a kinetic energy of 0J and a potential energy given by gravity

So we have that at his highest point, John's energy is given by:

Where  is the height above the ground.

Therefore, by conservation of energy, we have:

Since total energy can be determined in this situation as:

In our case,

Here,  is the mass of the object and  is its velocity

Here,   is the mass of the object,  is the gravitational constant, and  is the distance from the datum (where the potential energy is considered to be zero). During the process of the ball falling down, the total energy of the system is conserved, so at , the ball will have the same total energy as it does at . 

At this height, the ball has , , , and 

Plug in known values and solve.

This problem is best solved using mechanical energy conservation. All of the cart's kinetic energy  will be converted into gravitational potential energy  at the instant the cart comes to rest. Set these equations equal to each other. We find that mass is a common term, and gets cancelled out.

Solve for height.

We know that the total energy lost to the environment is equal to the negative change in mechanical energy. This is because all of the energy lost to the environment came from the decrease in mechanical energy. 

Change in mechanical energy  is given as:

Where  is change in kinetic energy and  is change in potential energy. 

First determine the change in kinetic energy:

Where  is the mass,  is the final velocity and  is the initial velocity. 

In this case, since the ball is going at the same speed initially and terminally , and has not changed its mass:

Now, let's determine the change in potential energy: 

Where  is the mass,  is the gravitational constant,  is the final height above the ground and  is the initial height above the ground. 

We initially started out 30m above the ground and came down to 25m above the ground. The mass is unchanged at 15kg, and the gravitational constant is given. Plug these values into the equation for change in potential energy:

Therefore,

Since 

There are three main quantities that suggest using conservation of energy: speed, height, and the compression of a spring. Reason being, these are the main terms that come up in the equations to describe these particular energies. 

Spring Potential Energy: 

Gravitational Potential Energy: 

Kinetic Energy: 

This principle we start from is the work-energy therorem that states the change in energy of a system is equal to the amount of non-conservative ork the system does on the enviornment.

Non conservative work typically comes in the form of friction. There are more types of non conservative forces, but they can become nebulous and they are often not dealt with in problems such as these. Considering friction is not a part of this problem we can simplify the above equation to:

This also implies the final energy is equal to the initial energy: 

Now we can write down every term that we deal with in the problem.

I am chosing the initial location to be when the spring is compressed and the ball is not in motion. The final location is where the ball is 1 meter from the ground. Eliminate terms that are zero:

GPE initial is zero because typically the zero of the problem is chosen at the lowest height(h=0). KE initial is zero because the ball is not in motion yet. SPE finial is zero because we assume the spring expanded back to it's equilibrium position after launching the ball. So we are left with: 

We can conceptually understand this equation as the energy held in the compressed spring has been transfered into displacing the ball upwards the the gravitiational field, and the energy also went it to giving the ball motion. Lets now show the variables at play:

We now can solve for the speed of the ball:

If you don't see how I solved for , ask someone for help with the algebra. Plugging into this equation we see the speed at 1 meters is .

The simplest way to solve this problem is to approach it from an energy perspective. We know that energy is always conserved, and in this case, air resistance is negligible. This means that we also have conservation of mechanical energy, since no energy will be lost to non-conservative frictional forces. Therefore, the object's gravitational potential energy will be completely converted into kinetic energy during its descent.

In equation form, we can write this as:

Or, put another way:

And since the difference between final and initial states is equal to the change in that state, we can further rewrite this as:

Since the object is initially starting at rest, we know that all of its mechanical energy is in the form of gravitational potential energy and therefore it doesn't have any kinetic energy. Thus, the value of its final kinetic energy will be equal to the change in its potential energy from  to .

To solve this problem, we must look at the total energy of the system at two different times: at the beginning before the motion starts, and at the end of the incline.  From there, we can use the conservation of energy to solve for the speed of the object at the bottom of the incline.