Question

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 $m/s^2$).

Answer 1

Taking upward direction as $+$

and the downward direction as $-$

with using $s=ut+\frac{1}{2}a{t}^2$

we need to put $u=10m/s$

$a=-g=-10m/s^2$ and $s=-40m$ $\text{(downward *displacement*)}$

we get the following equation, $-40=10t-5t^2$

or $t^2-2t-8=0$ or $t=4second$

Just employ the Shridharacharya Formula $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $t$ is root of $a{t}^2+bt+c=0$