Question

A ball is thrown with a velocity of $20m/sec$ from top of a multistory building the height of point from were the ball is thrown in $25m$ from the ground. (i) How height is the ball rise (ii) How long will it take before hitting the ground? Take $g=10m/s$

Answer 1

(1) for solving how high will ball go , velocity at the top point will be 0

this gives , $s=\frac{v^2-u^2}{2a}=\frac{0-{20}^2}{-2\times 10}$

$=20m$

(2)ball will go $25+20=45m$ high from the ground

for finding time , we have displacement $s=-25m,u=20m{s}^{-1},a=-10m{s}^{-2}$

using first equation, $s=ut+1/2a{t}^2$

we get $-25=20t-5t^2$

$\Rightarrow 5t^2-20t-25=0$

$\Rightarrow t^2-4t-5=0$

$\Rightarrow (t-5)(t+1)=0$

this gives$t=5secort=-1sec$

but we know that t cannot be negative so $t=5sec$