A ball whose kinetic energy is E, is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be (a) E (b) E/√(2) (c) E/2 (d) zero

Answer: (c). At the highest point the speed is ucos θ, so Ktop = Ecos ²θ. Ktop = Ecos ² 45° = E/2 So the answer is (c) E/2.

A ball whose kinetic energy is E, is projected at an angle of 45° to… | JEE Main 2002 (AIEEE)

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Conservation of Energy

A ball whose kinetic energy is E, is projected at an angle of $45^\circ$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

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Answer: (c)

At the highest point the speed is $u\cos\theta$ , so $K_{top} = E\cos^2\theta$ .
$K_{top} = E\cos^2 45^\circ = \frac{E}{2}$
So the answer is (c) $\frac{E}{2}$ .

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