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A body freely falls from a certain height onto the ground in a time ‘t’. During the first one third of the interval it gains a kinetic energy ${Δk}_{1},$ and during the last one third of the interval, it gains a kinetic energy ${Δk}_{2},$ The ratio ${Δk}_{1} : {Δk}_{2}$ is

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1 : 1

1 : 3

1 : 4

1 : 5

answer is D.

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Detailed Solution

downward distance covered in time t , $h = \frac{1}{2} {gt}^{2}$

applying Conservation of mechanical energy,

$∆ K \propto h$

$\Rightarrow ∆ K_{1} \propto \frac{t^{2}}{9}$

and, $∆ K_{2} \propto (t^{2} - \frac{4 t^{2}}{9}) = \frac{5 t^{2}}{9}$

$∴ {ΔK}_{1} : {ΔK}_{2} = 1 : 5$

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