Question

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 22.0 m below the point where the rock left your hand? Ignore air resistance.

Answer 1

Answer:

Explanation:

for vertical movement , time to reach the top = time to reach the hand = 2.5 s

v = u - gt

At the top , v = 0 , time t = 2.5 s

0 = u - g x 2.5

u = 2.5 x 9.8 = 24.5 m /s

velocity of throw = 24.5 m /s

So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v

v² = u² + 2 g s

= 24.5² + 2 x 9.8 x 22

= 600.25 + 431.2

= 1031.45

v = 32.11 m /s .