A particle is projected at600 to the horizontal with an energy E. The kinetic energy and potential energy at the highest point are
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a
(E/2, E/2)
b
(3E4, E4)
c
(E,0)
d
(E4, 3E4)
answer is D.
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Detailed Solution
Initial K.E., E=12mu2 At the highest point, velocity υ= u cos 600=u2 ∴ K.E. at highest point=12mυ2=12m(u2)2 =14(12mu2)=E4 PE at highest point =mg×u2sin26002g =12mu2(3 /2)2=34E .