First slide
Release from certain height
Question

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)

Easy
Solution

For both the bodies, displacement is same. From the given condition
2hgA2hgB=2 .......(1)  and 2gBh2gAh=10 ......(2)
dividing (2) by (1) and then solving
gAgB = 25

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