physics word problem

Question

A small catapult is used to launch 1.5 kg water balloons straight up into the air. A clever student is able to measure the speed of the water balloon while it is rising up into the air. She finds that the water balloon is 8.0 m above the catapult it is moving at a speed of 20.0 m/s. (For the purposes of this question, assume that the catapult is positioned in a small hole so that the height of the catapult is level with the ground.)

a) What is the total energy of the water balloon at a height of 8.0 m?

b) What is the maximum height that the water balloon will reach?

Answer 1

A small catapult is used to launch 1.5 kg water balloons straight up into the air. A clever student is able to measure the speed of the water balloon while it is rising up into the air. She finds that the water balloon is 8.0 m above the catapult it is moving at a speed of 20.0 m/s. (For the purposes of this question, assume that the catapult is positioned in a small hole so that the height of the catapult is level with the ground.)

a) What is the total energy of the water balloon at a height of 8.0 m?

total energy = kinetic energy + potential energy or

E = ½ m v^2 + m g h 

m = 1.5 kg v = 20.0 m/s g = 9.81 m/s^2 h = 8.0 m

E = ½ (1.5 kg) (20.0 m/s)^2 + (1.5 kg)(9.81 m/s^2)(8.0 m) = 420 J

b) What is the maximum height that the water balloon will reach?

By conservation of energy, the energy at the maximum height will be all potential (v’=0) and equal to the energy calculated in (a). Thus

m g H = E or H = E/mg = (½ m v^2 + m g h)/mg = ½ v^2/g + h

H = ½ (20.0 m/s)^2 / (9.81 m/s^2) + 8.0 m = 28 m

Let me know if I can explain any parts of these short answers more clearly. Thank you!