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 ?

Answer

velocity `=20 m s^(-1)`; time =2 s

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'
Initial Velocity of ball, u = 0
Distance or height of fall, s = 20 m
Downward acceleration, a = 10 m `s^(-2)`
As we know, 2as = `v^(2)-u^(2)`
`v^(2)=2as+u^(2)`
= `2xx10xx20+0`
= 400
`therefore` Final velocity of ball, v = 20 m`s^(-1)`
t = (v-u)/a
`therefore` Time taken by the ball to strike = (20-0)/10
=20/10
= 2 seconds