Viday’s classmate drops a ball from the top of the building. She hears the sound of the ball hitting the ground 2.057 seconds after dropping it. If she was 19.6 meters above the ground, what is the speed of sound in air (Take g = 9.8m/s2).

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Solution

Sol: As per given in the question that;
$h e i g h t : h = s = 19.6 m a c c e l e r a t i o n a = 9.8 m / s^{2} i n i t i a l v e l o c i t y u = 0 t i m e t a k e n t o r e a c h t h e b a l l a t g r o u n g t_{1} = ? S e c o n d e q u a t i o n o f m o t i o n : s = u t + \frac{1}{2} a t^{2} p u t t i n g v a l u e s : 19.6 = 0 \times t_{1} + \frac{1}{2} \times 9.8 \times {t_{1}}^{2} {t_{1}}^{2} = 4 t_{1} = 2 s e c A s p e r g i v e n i n t h e q u e s t i o n : t o t a l t i m e t a k e n : t = t_{1} + t_{2} = 2.057 s e c [t_{2} = t i m e t a k e n t o t r a v e l t h e s o u n d f r o m g r o u n d t o t h e t o p] s o t i m e t a k e n t o t r a v e l t h e s o u n d w i l l b e t_{2} = 2.057 - 2 = 0.057 s e c f o r m u l a : s p e e d = \frac{d i s \tan c e}{t i m e} = \frac{19.6}{0.057} = 343.85 m / s : A n s$

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