A boy on a cliff 480 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed does he throw the second stone? (Take g=10m/s²).

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Solution

$A s w e k n o w,$
$b y u s i n g 2 n d e q u a t i o n o f m o t i o n f o r t h e f i r s t s t o n e$
$S = u t + \frac{1}{2} a t^{2}$
$480 = 0 + \frac{1}{2} a t^{2}$
$t = 9.8 s e c$

$u s i n g t h i s t i m e i n t e r v a l f o r t h e s e c o n d s t o n e a s i t w i l l b e t a k i n g 1 s e c l e s s t i m e t h a n t h e p r e v i o u s s t o n e,$ t=8.8 sec
$b y u s i n g 2 n d e q u a t i o n o f m o t i o n f o r t h e 2 n d s t o n e$
$S = u t + \frac{1}{2} a t^{2}$
$480 = u (8.8) + \frac{1}{2} (10) {(8.8)}^{2}$
$u = 10.54 \frac{m}{s e c}$

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A boy on a cliff 480 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed does he throw the second stone? (Take g=10m/s2).

Asweknow,byusing2ndequationofmotionforthefirststone S=ut+12at2480=0+12at2t=9.8secusingthistimeintervalforthesecondstoneasitwillbetaking1seclesstimethanthepreviousstone,t=8.8 secbyusing2ndequationofmotionforthe2ndstone S=ut+12at2480=u(8.8)+12(10)(8.8)2u=10.54msec

A boy on a cliff 480 m high drops a stone. One second later he throws a second stone after the first. They both hit the ground at the same time. With what speed does he throw the second stone? (Take g=10m/s²).