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someone throws a rubber ball vertically upward from the roof of a building 8.2m in height. the ball rises, then falls.it just misses the edge of the roof, and strikes the ground.if the ball is in the air for 6.2s, what was its initial velocity?(disregard air resistance. a=-g=-10m/s²

User Ungalnanban
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1 Answer

11 votes
11 votes

Answer:

H = V t - 1/2 g t^2 V and g are in different directions

-8.2 = V * 6.2 - 1/2 * 10 * 6.2*2 = 6.2 V - 192.2

V = 184 / 6.2 = 29.7 m/s

Check: find time for ball to return to zero height

0 = 29.7 T - 5 T^2

T = 29.7 / 5 = 5.94 sec

The ball must have fallen 8.2 m in (6.2 - 5.94) sec = .26 sec

S = 29.7 * .26 + 5 * .26^2 = 8.1

User AshHimself
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