50 Points!
Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about the amounts of potential and kinetic energy the bowling ball has:
• as sits on top of a building that is 40 meters tall.
• as it is half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second.
• as it is just about to hit the ground from a fall off a building that is 40 meters tall and traveling 28 meters per second.

1. Does the bowling ball have more potential energy or kinetic energy as it sit on top of the building? Why?
2. Does the bowling ball have more potential energy or kinetic energy as it is half way through its fall? Why?
3. Does the bowling ball have more potential energy or kinetic energy just before it hits the ground? Why?
4. What is the potential energy of the bowling ball as it sits on top of the building?
5. What is the potential energy of the ball as it is half way through the fall, 20 meters high?
6. What is the kinetic energy of the ball as it is half way through the fall?
7. What is the kinetic energy of the ball just before it hits the ground?

1. It has more potential energy. Before the fall starts all the energy of the ball is potential energy, that is because its velocity is zero an its height is maximum.

2. As the ball falls all potential energy lost is equivalent to kinetic energy gain; when the ball is half way through its fall, both PE and KE are equal. That is, at the beginning all is potential energy, as ball falls PE starts to decrease, but KE starts to increase, so that, the addition of both equal the initial potential energy.

3. It has more kinetic energy. Just before the ball hits the ground, its height is near zero, then its PE is neglected, therefore, all the initial PE is now KE.

4. PE = m*g*h = 2*9.81*40 = 784 J

5. PE = m*g*h = 2*9.81*20 = 392 J

6. KE = 1/2*m*v^2 = 1/2*2*(19.8)^2 = 392 J

7. KE = 1/2*m*v^2 = 1/2*2*(28)^2 = 784 J