Question

A 2.0 kg rock is thrown from the top of a hill 60.0 m above sea level (the ravine) using a

catapult. The rock leaves the catapult with a speed of 10.0 m/s. The rock lands on the
ground that is 40.0 meters below sea level. Calculate the speed of therock just before it
contacts the ground (the ground is base level). Hint: use the conservation of mechanical
energy equation.

Answer 1

1/2 M V0^2 = initial KE of rock

KEi = 2 / 2 ^ 10^2 = 100 J

Gain in height of rock = 60 + 40 = 100 m = H (below starting point)

Gain in potential of rock = M g H = 2 * 9.8 * 100 = 1960 J

Total KE of rock = (100 + 1960) J = 2060 J

1/2 M V^2 = 2060 when rock lands

V^2 = 2060 m^2 / s^2

V = 45.4 m/s