Question

A 55 kg lead ball is dropped off the top of the Leaning Tower of Pisa. If the initial velocity of the ball is 0 m/s, what is the speed of the ball in m/s after falling 33.9 m?

Answer 1

Initial speed of the lead ball, u = 0

Distance covered by the ball in the downward direction, s = 33.9 m

Suppose speed of the lead ball after covering this distance of 33.9 m is 'v' m/s.

During free-fall, gravitational acceleration, g = 9.8 m/s^2 will act on the ball.

Use the expression -

v^2 = u^2 + 2*a*s

put the values -

v^2 = 0 + 2*9.8*33.9 [here, a = g = 9.8 m/s^2]

=> v = sqrt(2*9.8*33.9) = 25.78 m/s

Hence,

Speed of the lead ball after falling a distance of 33.9 m = 25.78 m/s (Answer)