A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of `10 ms^(-2),` with what velocity will it strike the ground ? After what time will it strike the ground?

Explanation

Here, `u =0,a = 10 ms ^(-2), s =20 m.`
`2as = v ^(2) - u^(2)`
`therefore 2 xx 10 ms ^(-2) xx 20 m = v ^(2) -0`
`therefore v ^(2) = 400 m ^(2) s ^(-2) therefroe v = 20 ms ^(-1)`
`v = u + at`
`therefore 20 ms ^(-1) = 0 + 1 0 ms ^(-2) xx t`
`therefore 20 ms ^(-1) = 10 ms ^(-2) xx t`
`therefore t = (20 ms ^(-1))/(10 ms ^(-2)) =2s `
The ball will strike the ground with a velocity of `20 ms ^(-1).` It will strike the ground after 2s.