Two balls are dropped from the same height at places A and B. The body at B takes two seconds less to reach the ground than the body at A and strikes the ground with a velocity greater than the body at A by 10 m/s. The product of the acceleration due to gravity at the two places A and B is ( in m2s–4)
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a
5
b
25
c
125
d
12.5
answer is B.
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Detailed Solution
For both the bodies, displacement is same. From the given condition2hgA−2hgB=2 .......(1) and 2gBh−2gAh=10 ......(2)dividing (2) by (1) and then solvinggAgB = 25