A ball is gently dropped from a height of 20 m. 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, u = 0
`a=10ms^(-2)`
`s=20m`
v=?
t=?
Using `v^2 -u^2 = 2as`, we get
`v^2-(0)^2=2xx10xx20`
`v^2=400`
`thereforev=sqrt(400)ms^(-2)=20ms^(-1)`
Now, using v = u + at, we get
`t=(vxxu)/a=(20-0)/10s=2s`