A body projected obliquely with velocity $19.6 {ms}^{- 1}$ has its kinetic energy at the maximum height equal to 3 times its potential energy there. Its position after 1second of projection from the ground is (h = maximum height)

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\frac{h}{3}

\frac{h}{2}

\frac{h}{4}

answer is D.

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Detailed Solution

K.E at max height

= \frac{1}{2} {mu}^{2} \cos^{2} θ

P.E at max height $= \frac{1}{2} {mu}^{2} \sin^{2} θ$

Given K.E = 3 P.E

$= \frac{1}{2} {mu}^{2} \cos^{2} θ = 3 \times \frac{1}{2} {mu}^{2} \sin^{2} θ$

$\tan^{2} θ = \frac{1}{3} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = 30 °$

$y = (u \sin θ) t - \frac{1}{2} {gt}^{2}$

$= 19.6 \times \frac{1}{2} - \frac{1}{2} \times 9.8 = 9.8 - 4.9 = 4.9 m$

$Max height = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{19.6}{4} = 4.9$

$y = h = 4.9 m$

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