A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

Explanation

According to the question, we draw the following figure

Given, `u=+10 m//s, a=-10 m//s^(2)` and `s=-40 m`
(at the point where stone strikes the gorund)
Substituting in `s=ut+(1)/(2)at^(2)`, we have
`-40=10t-5t^(2)` or `5t^(2)-10t-40=0` or `t^(2)-2t-8=0`
Sloving this we have t = 4s and `-2s`. Taking the positive value t = 4s.
Note : The significance of `t = -2 s` can be understood by following figure