Question

QUESTION 4 A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the approximate height from which it is dropped.​

Answer 1

S = V t + 1/2 g t^2 where the ball has a speed of V and falls for .2 sec

6 = V t + 4.9 * .04 = V * .2 + .2

V = (6 - .2) / .2 = 29 m/s speed entering area

T = V / g = 29 / 9.8

T = 2.96 sec time to reach speed V

H = 1/2 g t^2 time to fall a distance H

H = 4.9 * 2.96^2 = 42.9 m height from which ball was dropped

Check using first equation:

S = 29 * .2 + 4.9 + .2^2 = 6 m