A stone is thrown vertically upwards with velocity 20m/s. at the same time , and 20m vertically above . a second stone is allowed to fall. After whst time and at what height do they collide ?. take g = 10m/s^2


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Question Number 6164 by sanusihammed last updated on 16/Jun/16

$A s t o n e i s t h r o w n v e r t i c a l l y u p w a r d s w i t h v e l o c i t y 20 m / s .$ $a t t h e s a m e t i m e, a n d 20 m v e r t i c a l l y a b o v e . a s e c o n d s t o n e i s$ $a l l o w e d t o f a l l . A f t e r w h s t t i m e a n d a t w h a t h e i g h t d o t h e y$ $c o l l i d e ? .$ $t a k e g = 10 m / s^{2}$
Commented by prakash jain last updated on 17/Jun/16

$Let us say stone which is falling travels$ $distance x . Then stone thrown upwards$ $will travel distance 20 - x .$ $20 - x = u \cdot t - \frac{1}{2} {g t}^{2} = 20 t - 5 t^{2} (1)$ $x = 0 \cdot t + \frac{1}{2} {g t}^{2} = 5 t^{2} (2)$ $s u b s t i t u t i n g v a l u e o f x f r o m 2 t o 1 .$ $20 - 5 t^{2} = 20 t - 5 t^{2}$ $t = 1 s$ $x = 5 m (f r o m t h e t o p)$
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