A body covers 20 meters in last second of its free fall motion. From what height the body fall ?

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Solution

$D i s \tan c e c o v e r e d i n l a s t (n t h) s e c o n d : D i s \tan c e c o v e r t e d i n l a s t s e c o n d = D i s \tan c e c o v e r e d i n n s e c o n d s - D i s \tan c e c o v e r e d i n (n - 1) s e c o n d s 20 = \frac{1}{2} g n^{2} - \frac{1}{2} g {(n - 1)}^{2} 20 = \frac{1}{2} g n^{2} - \frac{1}{2} g n^{2} - \frac{1}{2} g + g n 20 = - \frac{1}{2} g + g n g n = 20 + \frac{g}{2} n = \frac{20}{g} + \frac{1}{2} = \frac{20}{9.8} + 0.5 = 2.54 s H e i g h t o f f a l l : S = \frac{1}{2} g n^{2} = \frac{1}{2} \times 9.8 \times 2 . 54^{2} = 31.6 m$

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A body covers 20 meters in last second of its free fall motion. From what height the body fall ?

Distance covered in last (n th) second:Distance coverted in last second=Distance covered in n seconds - Distance covered in (n-1) seconds20=12gn2-12gn-1220=12gn2-12gn2-12g+gn20=-12g+gngn=20+g2 n=20g+12 =209.8+0.5 =2.54 sHeight of fall:S=12gn2 =12×9.8×2.542 =31.6 m