# A stone is dropped into a river from a bridge 46.5 m above the water. Another stone is thrown vertically down 1.28 s after the first is dropped. Both stones strike the water at the same time. What is the initial speed of the second stone ?

Mar 3, 2018

$17 m {s}^{-} 1$

#### Explanation:

Suppose, the $2$ nd stone was thrown with a velocity of $u$ downwards,if it took time $t$ to reach the height of the bridge,then we can write,

$46.5 = u t + \frac{1}{2} g {t}^{2}$ ....1 (using, $s = u t + \frac{1}{2} g {t}^{2}$)

now,if the 1st stone after dropping(i.e it had no initial velocity) took time $t '$ to reach the height of the bridge then,

$46.5 = \frac{1}{2} g t {'}^{2}$

Given, $t ' - t = 1.28$

So,comparing the above three equations we can write,

$u t + \frac{1}{2} g {t}^{2} = \frac{1}{2} g {\left(t + 1.28\right)}^{2}$.....2

now,solving equation 1 and 2 we get, $u = 17 m {s}^{-} 1$