# A ball is projected with kinetic energy E at an angle of 45? to the horizontal. At the highest point during its flight, its kinetic energy will be

The expression for kinetic energy is KE= 1/2 mv

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When a body is projected with kinetic energy E and angle $\theta $ in a projectile motion, then at its highest point, the body has no vertical velocity ($v\mathrm{sin}\theta =0$) but it has the horizontal velocity (${v}_{II}=v\mathrm{cos}\theta $).

Thus, the kinetic energy of the body at its highest point will be,

${E}_{h}=\frac{1}{2}m{\left(v\mathrm{cos}\theta \right)}^{2}\phantom{\rule{0ex}{0ex}}{E}_{h}=\frac{1}{2}m{\left(v\mathrm{cos}45\right)}^{2}=\frac{1}{2}m{v}^{2}{\left(\frac{1}{\sqrt{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}{E}_{h}=\frac{E}{2}\phantom{\rule{0ex}{0ex}}so,atthehighestpointthekineticenergyis\frac{E}{2}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Regards

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