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It takes 1.8s to drop a ball of a bridge. How high is the bridge and what speed is reached?

t= 1.8s

a=-9.81ms^-²

v=u+at

v= -9.81 x 1.8 =17.658 m/s

s= ut + 1/2 at²(as u is 0 s= 1/2 at²)

s = 1/2 x 9.81 x 1.8²

Answered by Gabriella L. • Physics tutor

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