Question

at the moment when a shot putter releases a 4kg shot the shot is 1.25 m above the ground and travelling at 7 m/s. it reaches a maxium height of 3.7 m above the ground and then falls to the ground. air resistance is negligible and the shot goes straight up and then comes back down

What was the potential energy of the shot as it left the hand relative to the ground?

What was the kinetic energy of the shot as it left the hand?

What was the total energy of the shot as it reached its maximum height?

What was the kinetic energy of the shot just as it struck the ground?

Answer 1

Take potential = 0 at ground level as instructed.

1. m g h = 4 * 9.81 * 1.25 =49.05 J

2. (1/2) m v^2 = .5*4*49 = 98 J

3. No friction so PE + Ke = constant
= 49.05 + 98 = 147.05 J
by the way that means when it stops at the top it should have potential energy of 147 J
147 = m g h = 4*9.81 * h
h = 3.74, well they are a little off, maybe took 9.8 or 10 for g instead of 9.81

4. Potential energy lower by 49.05 because the ground is lower so it speeds up. All the energy is kinetic now
17.05

Answer 2

typo
147.05 J for #4