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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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