# An object is thrown vertically from a height of 9 m at  5 m/s. How long will it take for the object to hit the ground?

Mar 11, 2018

Let's construct an equation for the motion of the object.

Well here we will be using, $S = u t + \frac{1}{2} a {t}^{2}$

Considering upward direction to be positive, we get, $S = - 9$ as on reaching the ground it will have a net displacement of $9 m$ downwards.

$u = 5 m {s}^{-} 1$ and $g = - 9.8 m {s}^{-} 2$

So,if it takes time $t$ to reach the ground we can write,

-9 = 5t -1/2 ×9.8×t^2

Solving this we get, $t = 1.96 s$