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

Explanation

Here, `" "` Initial velocity, `u` = 0 `" "` (Ball dropped from rest)
`" "` Final velocity, `v` = ? `" "` (To be calculated)
`" "` Acceleration, `a` = 10 m `s^(-2)`
And, `" "` Distance (or Height), `s` = 20 m
Now, `" "v^(2)=u^(2)+2as`
`" "v^(2)= (0)^(2)+2xx10xx20`
`" "v^(2 )=0 +400`
`" "v^(2)= 400`
`" "v=sqrt(400)`
`" "v=20" m "s^(-1)`
Thus, the ball will strike the ground with a velocity of 20 metres per second. Let us calculate the time now.
We know that : `" "v=u +at`
So, `" "20=0+10xxt`
`" "10t= 20`
`" "t = (20)/(10)`
`" "t= 2 "s"`
Thus, the ball will strike the ground after 2 seconds.