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A particle is projected at 60∘ to the horizontal with kinetic energy K. The kinetic energy at the position when particle makes 30∘with horizontal is ___K

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answer is 0.33.

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Detailed Solution

K=12mu2tan⁡α=vyvx⇒tan⁡30=vyvx⇒vy=vx3KEf=12mvx2+vy2=12mux2+vx32KEf=12m43u2cos2⁡60=k×43×14=0.33K

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