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  • A ball is projected with kinetic energy E, at an angle of to the horizontal. The kinetic energy of

A ball is projected with kinetic energy E, at an angle of 60^{\circ} to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Option: 1

Zero


Option: 2

\mathrm{\frac{E}{2}}


Option: 3

\mathrm{\frac{E}{4}}


Option: 4

\mathrm{E}


Answers (1)

best_answer

\mathrm{u_{x}=u\cos 60\degree=\frac{u}{2}}

\mathrm{u_{y}=u\cos 60\degree=\frac{\sqrt{3}u}{2}}

\mathrm{E=\frac{1}{2}mu^{2}}

At the highest point ,

\mathrm{Kinetic \: energy=\frac{1}{2}mv^{2}}

                            \mathrm{=\frac{1}{2}m\left ( u\cos 60\degree \right )^{2}}

                            \mathrm{=\frac{1}{4}\left ( \frac{1}{2}mu^{2} \right )}

                            \mathrm{=\frac{1}{4}\left ( E \right )=\frac{E}{4}}

Hence 3 is correct option

Posted by

Gautam harsolia

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