### Video Transcript

A cat’s crinkle ball toy of mass 15 grams is thrown straight up with an initial speed of 3.0 meters per second. Assume in this problem that air drag is negligible. What is the kinetic energy of the ball as it leaves the hand? How much work is done by the gravitational force during the ball’s rise to its peak? What is the change in gravitational potential energy of the ball during the rise to its peak? If the gravitational potential energy is taken to be zero at the point where the ball leaves the hand, what is the gravitational potential energy when it reaches the maximum height? If the gravitational potential energy is taken to be zero at the maximum height the ball reaches, what would the gravitational potential energy be when it leaves the hand? What is the maximum height the ball reaches?

Let’s begin this multipart problem by recording information we’ve been given. We’re told the mass of the ball is 15 grams; we’ll call that 𝑚. And we’re told that the ball is thrown upward from the hand with an initial speed of 3.0 meters per second; we’ll call that 𝑣 sub 𝑖. We want to know, in order, the kinetic energy of the ball just as it leaves the hand; we’ll call that KE. Then we want to know the work done by gravity on the ball as it rises; we’ll call that 𝑊 sub 𝑔. Next, we wanna find out the change in the ball’s gravitational potential energy over the course of its ascent; we’ll call that ΔPE. Next, we wanna know the gravitational potential energy of the ball at its maximum height if we set the gravitational potential energy to zero at the position of the person’s hand; we’ll call this PE sub 𝑡, for the potential energy at the top of the ball’s path.

Then if we move the zero point of gravitational potential energy to the apex of the ball’s trajectory, we want to solve for its gravitational potential energy just as it leaves the hand; we’ll call this PE sub 𝑏, for potential energy at the bottom of its flight. Finally, we wanna know the maximum height the ball reaches; we’ll call this ℎ sub max. Let’s begin to solve for the kinetic energy of the ball as it leaves the hand by recalling the relationship for kinetic energy. The kinetic energy of an object is equal to one-half its mass times its speed squared. In our case, the mass is 15 grams and the speed is 3.0 meters per second. Before we insert these values in the equation, let’s convert our mass from units of grams to units of kilograms. 1000 grams is equal to one kilogram, so 15 grams is equal to 0.015 kilograms.

We then insert the value for 𝑣 sub 𝑖. And when we calculate this value, we find the ball has a kinetic energy of 0.068 joules. This is the ball’s energy due to its motion just as it leaves the hand. Now let’s move on to solve for the work done by gravity on the ball as it rises. If we draw a sketch of the ball’s ascent, we can see that the motion of the ball as it rises opposes the direction of the force of gravity on the ball. To solve for 𝑊 sub 𝑔, we can use the work-energy theorem. This theorem connects the work done on an object with its change in kinetic energy. Considering the ball tossed into the air and coming to a stop, we can see that its initial kinetic energy is 0.068 joules and its final kinetic energy, the kinetic energy at the top of the ball’s flight, is zero because at that point the ball is not moving and therefore has zero velocity.

Therefore, the change in kinetic energy of the ball over its ascent, ΔKE, is equal to the final kinetic energy minus the initial kinetic energy of the ball. This is equal to negative 0.068 joules. And by the work-energy theorem, this is equal to the work done by gravity on the ball as it rises. Notice that the minus sign makes sense because gravity acts in a downward direction through its force, yet the ball’s displacement is upward as it ascends. Now we can move on to solving for the change in gravitational potential energy of the ball as it rises. As we approach this question, let’s recall the law of conservation of energy. In its simplest form, this law tells us that the initial energy in a system is equal to the final energy within that closed system.

In the case of the ascending ball, if we add up the kinetic and potential energy of the ball as soon as it leaves the hand, that sum will equal the sum of the kinetic and potential energy of the ball when it’s at its highest point, ℎ max. At that point, ℎ max, we know the ball is not in motion so it’s kinetic energy is zero. And just as the ball leaves the hand, we can set that point as the zero point for potential energy, so that term is zero on the left side. This means that the change in the ball’s potential energy, ΔPE, during its ascent is equal simply to the final potential energy of the ball. And by the conservation of energy, that is equal to the initial kinetic energy of the ball.

We solved for that value in the first part of this problem. KE sub 𝑖 is equal to 0.068 joules, which is equal to ΔPE, so the ball’s change in potential energy over its ascent is equal to its initial kinetic energy as soon as it leaves the hand. Now we move on to solving for the gravitational potential energy of the ball when it’s at the top of its path. We’re told to assume that for this question, the gravitational potential energy is set to zero at the point where the ball initially leaves the hand. With this constraint, we know that PE sub 𝑡, the ball’s gravitational potential energy at its highest point, is equal to ΔPE, the value we just solved for, because we made the same assumption for ΔPE about the location of the zero point of gravitational potential energy.

Therefore, PE sub 𝑡 is equal to 0.068 joules. Now we move on to solving for the gravitational potential energy of the ball at its lowest point, just as it leaves the hand. In this case, instead of setting the zero point of gravitational potential energy to be where the ball leaves the hand, we set it instead at the ball’s highest point, ℎ max. As we solve for PE sub 𝑏, we can rely on a previously obtained answer to help us. PE sub 𝑏 is equal to the gravitational potential energy of the ball at its maximum height, which is zero, minus the change in potential energy of the ball as it goes from the hand to its maximum height, ΔPE. We’ve solved for ΔPE and found it is 0.068 joules, therefore PE sub 𝑏 is equal simply to negative 0.068 joules.

This result depends on two things: where we set gravitational potential energy to be zero and the change in gravitational potential energy of the ball over its flight. Finally, we want to solve for ℎ max, the maximum height that the ball obtains. To do this, let’s recall the relationship for gravitational potential energy. An object’s gravitational potential energy equals its mass times 𝑔, the acceleration due to gravity, times its height. Applying this relationship to our situation, PE sub 𝑡, the potential energy of the ball at its maximum height, is equal to the ball’s mass times the acceleration due to gravity times ℎ max.

If we divide both sides of the equation by 𝑚 times 𝑔, those two terms cancel out on the right side of our equation, leaving us with an equation for ℎ max, that it’s equal to PE sub 𝑡 divided by 𝑚𝑔. We’ve solved for PE sub 𝑡, that’s 0.068 joules. And 𝑚, the mass of the ball in kilograms, is 0.015 kilograms. And 𝑔, the acceleration due to gravity, we’ll treat as exactly 9.8 meters per second squared. When we compute this value for ℎ max, we find that it’s a height of 46 centimeters. This is how high above its release point that the ball will rise.