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

Question

Dr. Abubacker · Accepted Answer

(c) At the highest point of its flight, vertical component of velocity is zero and only horizontal component is left which is
$u_{x} = u\cos	heta$
Given $	heta = 45{^\circ}$
$	herefore\ \ u_{x} = u\cos 45{^\circ} = \frac{u}{\sqrt{2}}$
Hence, at the highest point kinetic energy
$E^{'} = \frac{1}{2}mu_{x}^{2}$
$= \frac{1}{2}m\left( \frac{u}{\sqrt{2}} ight)^{2} = \frac{1}{2}m\left( \frac{u^{2}}{2} ight)$
$= \frac{E}{2}\ \ \left( \because\frac{1}{2}mu^{2} = E ight)$

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

Explanation

Reviewed by Dr. Abubacker

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