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

Explanation

Let us assume, the final velocity with which ball will strike the ground be .v. and time it takes to strike the ground be .t.
Initital velocity of ball, u = 0
Distance or height of falls, S= 20m
Downward acceleration `a = 10 ms^(-2)`
As we know, `2as = v^(2) - u^(2)`
`v^(2) = 2as + u^(2)`
`=2 xx 10 xx 20 + 0`
=400
Therefore Final velocity of ball,
`v= 20ms^(-1), t= (v-u)//a`
Therefore Time taken by the ball of strike `(20-0)//10`
`=20//10`
=2 seconds