Standing at the edge of a cliff 36.0 m high, you drop a ball. Then you throw a 2nd ball downward w/an?

Question

Standing at the edge of a cliff 36.0 m high, you drop a ball. Later, you throw a second ball downward with an initial speed of 8.0 m/s.
(a) The speeds of both balls increase when they reach the base of the cliff. Is the increase of the first ball's speed greater than, less than, or the same as that of the second ball?
greater than
less than
the same as



(b) Verify your answer to part (a) with a calculation. (v) of each.
v = ? m/s (first ball)
v = ? m/s (second ball)

Answer 1

(a) less than
(b) First ball is a free-falling object. Thus:
v = sqrt(2gh) = sqrt(2*9.8*36) = 26.6(m/s) (first ball)
The second ball should use a little sophisticated formula:
v^2 = (vo)^2 + 2gh
= 64 + 2*9.8*36
= 769.6
Hence v = 27.7(m/s) (second ball)