A boy, standing on a 49 m high cliff drops a stone. One second later he throws another stone vertically downwards. Both the stones hit the ground at the same time. With what speed was the second stone thrown?

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Solution

We have to find $v_{o}$ , the speed with which he throws the second ball.
For the first ball, u = 0
s = 49 m
a = 9.8 $m / s^{2}$
t = ?
From the second equation of motion, $s = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 49 = \frac{1}{2} 9.8 \times t^{2}$
$\Rightarrow t^{2} = 10 \Rightarrow t = \sqrt{10} = 3.16 s$
For the second ball, $u = v_{o}$
s = 49 m
a = 9.8 $m / s^{2}$
t = 3.16 - 1 = 2.16 s
$s = v_{o} t + \frac{1}{2} a t^{2}$
$\Rightarrow 49 = v_{o} \times 2.16 + \frac{1}{2} \times 9.8 \times 2.16 \times 2.16$
$\Rightarrow v_{o} = 12.1 m / s$

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