Question

A boy on a 49m high drops a stone once again later he throw another stone.Both the stone hit at the same time.what speed if the second stone thrown.​

Answer 1

$\sf\star\bold\purple{{QUESTION:-}}$

✓✓A boy on 49m high drops a stone. one second later, he throw another stone.Both the stone hit at the same time.what speed if the second stone thrown.

$\sf\star\bold\green{{SOLUTION:-}}$

$\sf\pink{{For\:First\:stone:-}}$

• $\sf\bold\:s=49\:m$

• $\sf\bold\:u=0m/s$

• $\sf\bold\:g=9.8m/{s}^{2}$

• $\sf\bold\:t=?$

$\sf\bold\red{{According\:to\: question:-}}$

$\sf\:s=ut+\dfrac{1}{2}g{t}^{2}$

$\sf\implies\:49=0\:\times\:t+\dfrac{1}{2}\times\:9.8\times\:{t}^{2}$

$\sf\implies\:49=\dfrac{1}{2}\times\:\dfrac{9.8}{10}\times\:{t}^{2}$

$\sf\implies\:49=4.9{t}^{2}$

$\sf\implies\:{t}^{2}=\dfrac{49}{4.9}$

$\sf\implies\:t={ \sqrt{10}}$

$\sf\implies\:t=3.16s$

$\sf\pink{{For\: second\:stone:-}}$

• $\sf\bold\:s=49\:m$

• $\sf\bold\:u=?$

• $\sf\bold\:g=9.8m/{s}^{2}$

• $\sf\bold\:t=(3.16-1)=2.16$

$\sf\bold\red{{According\:to\: question:-}}$

$\sf\:s=ut+\dfrac{1}{2}g{t}^{2}$

$\sf\implies\:49=u\times\:2.16+\dfrac{1}{2}\times\:9.8\times\:({2.16})^{2}$

$\sf\implies\:49=2.16u+4.9\times\:4.66$

$\sf\implies\:49=2.16u+22.83$

$\sf\implies\:2.16u=49-22.83$

$\sf\implies\:2.16u=26.17$

$\sf\implies\:u=\dfrac{26.17}{2.16}$

$\sf\implies\:u=\dfrac{2617}{216}$

$\sf\green{{\implies\:u=12.11m/s}}$