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Grade: upto college level
        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.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							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.

						
5 years ago
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