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 and assuming it to be uniform.

Answer

D

Explanation

consider the motion of ball from A to B.`Brarr ` just above the sand (just to penetrate) `u=0, a=9.8 m/s^2
S=5m`
From `S= ut+1/2at^2
rarr 5=0+1/2(9.8)t^2
rarr t^2=5/4.9= 1.02
rarr t= 1.01
:. Velocity at B
v'= u+at
= 9.8xx1.01=9.89m/s`
……..[Since u=0], Form motion of ball in sand
`u_1 = 9.89 m/s
v_1 =0
S=10 cm = 0.1 m.
S_2 =(v_1^2-u_1^2)/(2a)
S_2 =(v_1^2-u_1^2)/(2a)
=(0-(9.89)^2)/(2xx0.1)=-490m/s^2`
Hence retardation is sand is 490 m/s^2.