Question

A ball whose kinetic energy is $$E$$, is thrown at an angle of $$45$$ with the horizontal, its kinetic energy at the highest point of its flight will be :

A
E B
E2 C
E2 D
Zero Solution

The correct option is D $$\displaystyle \frac {E}{2}$$Let $$\theta =45^o$$ is the angle made by the ball with the horizontal.At highest point there is only horizontal velocity $${ v }_{ h }=vcos\theta \quad =v\cos45^o=\dfrac { v }{ \sqrt { 2 } }$$Kinetic energy  $$E=\dfrac { m{ v }^{ 2 } }{ 2 } \quad ;\quad E\propto { v }^{ 2 }$$if velocity decrease by a factor of $$\sqrt 2$$ then kinetic energy will decrease by a factor of 2.Kinetic energy at the highest point of its flight will be  $$\dfrac E2$$.Physics

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