A ball is thrown up with a speed of `15 m//s`. how high will it go before it begins to fall ?
Take `g=9.8 m//s^(2)`.

Answer

`11.48 M`

Explanation

Here, `u=15 m//s, v=0, g=-9.8 m//s^(2) h=?`
From `v^(2)-u^(2)=2gh,h=(v^(2)-u^(2))/(2g)=(0-15^(2))/(2(-9.8))=11.48m`