A ball falls freely from rest. The ratio of the distances travelled in the first, second, third, fourth second is?
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Solution

Distance travel is time t is :
$s = u t + 1 / 2 a t^{2} i n i t i a l v e l o c i t y u = 0 s = 1 / 2 a t^{2} d i s \tan c e t r a v e l l e d i n 1 s e c o n d = s_{1} = 1 / 2 \times 9.8 \times 1^{2} = 4.9 m s i m i l a r l y d i s t n a c e t r a v e l l e d i n 2, 3, 4 s e c o n d s a r e s_{2} = 19.6 m s_{3} = 44.1 m s_{3} = 78.4 m$

Now distance travelled in first(D₁), second(D₂) , third(D₃) and fourth(D₄) second are:

$D_{1} = s_{1} = 4.9 m D_{2} = s_{2} - s_{1} = 19.6 - 4.9 = 14.7 m D_{3} = s_{3} - s_{2} = 44.1 - 19.6 = 24.5 m D_{4} = s_{4} - s_{3} = 78.4 - 44.1 = 34.3 m r a t i o D_{1 :} D_{2} : D_{3} : D_{4} 1 : 3 : 5 : 7$

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