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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
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
