Dropping a Ball - Earth/Moon It takes 2.1 sec for a small ball released from res

Question

Dropping a Ball - Earth/Moon

It takes 2.1 sec for a small ball released from rest from a tall building to reach the ground. The mass of the ball is 0.04 kg. Calculate the height (in m) from which the ball is released. (Neglect air resistance.)


If that ball had been released from the same height, but this time, near the surface of the moon, how long (in seconds) would it have taken for the ball to hit the surface of the moon? The magnitude of the acceleration of gravity near the moon surface is 1.67 m/s2.

Explanation / Answer

given that the time taken by the ball to reach the ground from height h is t = 2.1s Mass of ball is m = 0.04 kg ---------------------------------------------------- For freely falling body initial velocity u = 0 just after the released Apply kinematics equation S = ut + (1/2)at^2 h = 0 + (1/2)gt^2 h = (1/2)(9.8 m/s^2)(2.1s)^2 h = 21.61 m --------------------------------------------------------------------------------------- Now the ball released from heigh h = 21.61 m on the moon, For freely falling body initial velocity u = 0 just after the released Apply kinematics equation S = ut + (1/2)at^2 h = 0 + (1/2)gt^2 21.61 = (1/2)(1.67 m/s^2)t^2 t^2 = 25.88 t = 5.087s (or) t = 5.09 s

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