A baseball is dropped off the side of a cliff. It free-falls to the ground in 8 seconds. During the third second, the ball is traveling about 38 m/s. a0

For this item since it was mention that the ball was freely dropped, the initial velocity should be zero. I assume that the unknown here is the distance traveled or the height of the cliff which can be calculated through the equation,

h = V₀t + 0.5gt²

where h is the height, V₀ is the initial velocity (in this case, zero), t is time, g is the acceleration due to gravity. Substituting the known values,

h = (0) (8) + 0.5 (9.8 m/s²) (8 s) ²

h = 313.6 m

Thus, the cliff is approximately 313.6 m high.