A body has kinetic energy $E$ when projected at angle of projection for maximum range. Its kinetic energy at the highest point of its path is:

E

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\frac{E}{2}

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\frac{E}{\sqrt{2}}

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Zero

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Solution

The correct option is D $\frac{E}{2}$
Maximum range is obtained at $45^{o}$ .

$E = \frac{1}{2} m u^{2}$ .....(i)

At highest point, $v = u cos 45^{o} = \frac{u}{\sqrt{2}}$

$∴ E^{'} = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{u}{\sqrt{2}})^{2}$

$= \frac{(\frac{1}{2} m u^{2})}{2} = \frac{E}{2}$

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A body has kinetic energy E when projected at angle of projection for maximum range. Its kinetic energy at the highest point of its path is:

The correct option is D E2Maximum range is obtained at 45o.E=12mu2.....(i)At highest point, v=ucos45o=u√2∴E′=12mv2=12m(u√2)2=(12mu2)2=E2

A body has kinetic energy $E$ when projected at angle of projection for maximum range. Its kinetic energy at the highest point of its path is:

The correct option is D $\frac{E}{2}$
Maximum range is obtained at $45^{o}$ .

$E = \frac{1}{2} m u^{2}$ .....(i)

At highest point, $v = u cos 45^{o} = \frac{u}{\sqrt{2}}$

$∴ E^{'} = \frac{1}{2} m v^{2} = \frac{1}{2} m (\frac{u}{\sqrt{2}})^{2}$

$= \frac{(\frac{1}{2} m u^{2})}{2} = \frac{E}{2}$