A boy on a cliff $49 m$ high, drops a stone, one second later, he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?

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Solution

Let time taken by thrown stone = $t$

Time taken by dropped stone = $t + 1$

s = $\frac{1}{2}$ g ${(t + 1)}^{2}$

substitute values of $g = 9.81$ and $s = 49 m$

t = $\sqrt{10}$ -1

s = $\frac{1}{2}$ g ${(t + 1)}^{2}$ = ut + $\frac{1}{2}$ g $t^{2}$

s = gt + $\frac{g}{2}$ = ut

u = g(1 + $\frac{1}{2}$ t)

= 12.08 $\frac{m}{s e c}$

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A boy on a cliff 49 m high, drops a stone, one second later, he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?

Let time taken by thrown stone = t Time taken by dropped stone = t+1 s = 12g(t+1)2 substitute values of g=9.81 and s=49 m t = √10-1 s = 12g(t+1)2 = ut + 12gt2 s = gt + g2 = ut u = g(1 + 12t) = 12.08 msec

A boy on a cliff $49 m$ high, drops a stone, one second later, he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?