From a cliff of $49 m$ high, a man drops a stone. One second later, he throws another stone. They both hit the ground at the same time. Find out the speed with which he threw the second stone.

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Solution

Step 1: Given
Height of the cliff, $s = 49 m$
Time taken by the first stone, $t s e c o n d s$
Time taken by the second stone, $(t - 1) s e c o n d s$
The initial velocity of stone one, $u_{1} = 0 m / s$
Acceleration due to gravity, $a = g = 9.8 m / s$
Step 2: Find the velocity
Let $v$ be the final velocity of the first stone,
From the equation of motion, on the first stone,
$v^{2} = {u_{1}}^{2} + 2 a s v^{2} = 0 + 2 (9.8) (49) v^{2} = 960.4 v = 30.99 v \approx 31 m / s$
Step 3: Find the time
Using the equation of motion on the first stone,
$v = u + a t 31 = 0 + 9.8 t \Rightarrow t = 3.16 s$
Therefore, for stone two,
$t - 1 = 3.16 - 1 = 2.16 s$
Step 4: Find the initial velocity of the second stone
Using the equation of motion, $s = u t + \frac{1}{2} a t^{2}$
For stone two,
$49 = u_{2} (t - 1) + \frac{1}{2} a {(t - 1)}^{2} 49 = u_{2} (2.16) + \frac{1}{2} (9.8) (2.16) \Rightarrow u_{2} = 12.1 m / s$
Hence, he threw the second stone with the speed of $12.1 m / s$ .

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