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Grade upto college level Mechanics

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Profile image of Amit Saxena
12 Years agoGrade upto college level
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
Sol. Consider the motion of ball form A to B. B → just above the sand (just to penetrate) u = 0, a = 9.8 m/s2, s = 5 m s = ut + ½ at2 ⇒ 5 = 0 + 1/2 (9.8)t2 ⇒ t2 = 5/4.9 = 1.02 ⇒ t = 1.01. ∴ velocity at B, v = u + at = 9.8 × 1.01 (u = 0) =9.89 m/s. From motion of ball in sand u1 = 9.89 m/s, v1 = 0, a = ?, s = 10 cm = 0.1 m. a = (V_1^(2 )- u_1^2)/2s = (0-( 9.89)^2)/(2 x 0.1) = -490 m/s2 The retardation in sand is 490 m/s2.

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