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2nd Equation of Motion

a stone is dr...

Question

a stone is dropped freely in the river frrom a bridge. It takes 5 sec to touch the water surface in the water. Calculate (a) The height of bridge from water level. (b) The distance covered by stone in the last second (g= 9.8 m/s2

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Solution

$D i s \tan c e t r a v e l l e d i n 5 s e c o n d s w i l l b e e q u a l t o t h e h i g h t o f t h e b r i d g e f r o m t h e w a t e r . S = u t + \frac{1}{2} a t^{2} A s s t o n e i s d r o p p e d, i n i t i a l v e l o c i t y i s z e r o, i n a f r e e f a l l a c c e l e r a t i o n i s e q u a l t o g . h = 0 \times t + \frac{1}{2} \times 9.8 \times 5 \times 5 h = 122 . . 5 m D i s \tan c e t r a v e l l e d i n l a s t s e c o n d = d i s \tan c e t r a v e l l e d i s 5 s - d i s \tan c e t r a v e l l e d i n 4 s . = 122.5 - (\frac{1}{2} \times 9.8 \times 4 \times 4) = 122.5 - 78.4 = 44.1 m$

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