A ball is dropped onto a sandy floor from a height of 5 m and penetrates the sand up to 10 cm before coming to rest. Assuming the retardation of the ball in the sand is uniform, determine its value in m/s ^2 .

Step-by-step solution

The velocity of the ball just before entering the sand is given by the equation of motion: v^2 = u^2 + 2gh Since the ball is dropped, its initial velocity u = 0 . Taking g = 9.8 m/s ^2 and h = 5 m : v^2 = 0 + 2 9.8 5 = 98 m ^2/ s ^2 Let a be the uniform retardation of the ball inside the sand. The final velocity v_f becomes zero after penetrating a distance s = 10 cm = 0.1 m . Using the third equation of motion for the motion inside the sand: v_f^2 = v^2 - 2as Substituting the values: 0 = 98 - 2 a 0.1 0.2a = 98 a =