# Question #9846e

Sep 10, 2017

The distance from the base of the tower that body will be found is
$v \cdot \sqrt{\frac{2 \cdot H}{9.8 \frac{m}{s} ^ 2}}$

#### Explanation:

The time to hit the ground is given by t in the equation
$s = u \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$
where s = H, u = 0, $a = g = \frac{9.8 m}{s} ^ 2$. Therefore
$t = \sqrt{\frac{2 \cdot H}{9.8 \frac{m}{s} ^ 2}}$

Let the velocity at which it was thrown be v. That means that in time t, the velocity v will have carried the body a horizontal distance
$v \cdot t = v \cdot \sqrt{\frac{2 \cdot H}{9.8 \frac{m}{s} ^ 2}}$

I hope this helps,
Steve