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Q.

A body projected obliquely with velocity 19.6 m/s 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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a

h/2

b

h/4

c

h/3

d

h

answer is D.

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Detailed Solution

KEat maximum height =3 PEat maximum height12mu2Cos2θ=3mgu2Sin2θ2gCos2θ=3Sin2θ hence θ=300y=uSinθt-12gt2 y=19.612×1-129.812=4.9 mh=u2Sin2θ2g=19.622×9.8×14=4.9mso y=h

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A body projected obliquely with velocity 19.6 m/s 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)