A boy on a cliff $49 m$ high drops a stone.One second later,he throws a second after the first .They both hit the ground at the same time.With what speed did he throw the second stone?

12.1 m / s

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10 m / s

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21.1 m / s

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15.1 m / s

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Solution

The correct option is A $12.1 m / s$
If time for first stone is $t$ , time for second stone is $t - 1$
$s = 49 m$ is the same for both stones, $u = 0$ for first stone
$v^{2} = u^{2} + 2 a s$
$v^{2} = 0 + 2 (9.8) (49)$
$v = 31 m / s$
$v = u + a t$
$31 = 0 + 9.8 t$
$t = 3.16 s$
$t^{'} = 2.16 s$
$s = u t + 0.5 a t^{2}$
$49 = u^{'} (2.16) + 0.5 (9.8) (2.16)^{2}$
$u^{'} = 12.1 m / s$

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A boy on a cliff 49m high drops a stone.One second later,he throws a second after the first .They both hit the ground at the same time.With what speed did he throw the second stone?

The correct option is A 12.1m/sIf time for first stone ist, time for second stone is t−1s=49m is the same for both stones, u=0 for first stonev2=u2+2asv2=0+2(9.8)(49)v=31m/sv=u+at31=0+9.8tt=3.16st′=2.16ss=ut+0.5at249=u′(2.16)+0.5(9.8)(2.16)2u′=12.1m/s

A boy on a cliff $49 m$ high drops a stone.One second later,he throws a second after the first .They both hit the ground at the same time.With what speed did he throw the second stone?