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A particle takes n seconds less and acqu...

A particle takes `n` seconds less and acquires a velocity u m/sec. higher at one place than at another place in falling through the same distance. Calculate the product of the acceleration due to gravity at these two places.

Text Solution

Verified by Experts

The correct Answer is:
`((u)/(n))^(2)`

`s=(1)/(2)g_(1)t_(0)^(2)` ..(i)
`s=(1)/(2)g_(2)(t_(0)-n)^(2)` …(ii)
`u_(0)=g_(1)t_(0)` ..(iii)
`u_(0)+u=g_(2)(t_(0)-n)` ..(iv)
After solving we get `g_(1)g_(2)=((u)/(n))^(2)`
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