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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h/4

h/3

h/2

answer is D.

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Detailed Solution

KE_{atmax height} =3 x PE_{atmax height}
$\begin{array}{l} \Rightarrow \frac{1}{2} m (ucos θ)^{2} = 3 (mg) [\frac{u^{2} \sin^{2} θ}{2 g}] \\ \Rightarrow \cos^{2} θ = 3 \sin^{2} θ \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = 30^{\circ} \end{array}$

$\begin{array}{l} y = usin θt - \frac{1}{2} {gt}^{2} \\ = 19.6 \times \frac{1}{2} \times 1 - \frac{1}{2} (6.8) (1)^{2} = 4.9 m \end{array}$

$And h = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{(19.6)^{2} \times \frac{1}{4}}{2 \times 9.8} = 4.9 m$
hence y = h

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