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A ball, having the total Kinetic energy as $E$ , is projected at an angle of $45 °$ with the horizontal. What would be the K.E. at the highest point of projection?

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0

E / 2

E / \sqrt 2

E

answer is B.

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Detailed Solution

A ball which is having the total Kinetic energy as E, is projected at an angle of

45 °

with the horizontal. The K.E. at the highest point of projection will be

\frac{E}{2} .

The kinetic energy of a body with mass

m

and velocity

V

is related by

E = \frac{1}{2} m V^{2}

The horizontal component of the velocity will be,

V_{1} = vcos 45 °

Here we need to find the K.E. at the highest point (

K E_{h}

) of projection KE,

K E_{h} = \frac{1}{2} m {(vcos 45 °)}^{2}

= \frac{1}{2} m v^{2} {(\cos 45 °)}^{2}

= E {(\cos 45 °)}^{2}

= E \times {(\frac{1}{\sqrt{2}})}^{2}

\Rightarrow K E_{h} = \frac{E}{2}

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