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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.​

Sagot :

Answer:

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

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