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A ball whose kinetic energy is E, is thrown at anangle of 45° with the horizontal, its kinetic energyat the highest point of its flight will be :-AIPMT 1997, 2000](4) zero2 |
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Answer» Given:let v be initial velocity.K.E= E =1/2 mV²VELOCITY OF PARTICLE AT HIGHEST POINT = vcos 45kinetic energy of particle at highest point is =1/2 m(vcos45)²=1/2(mv²)(1/2)=1/2 E |
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