A ball is projected with kinetic energy E , at an angle of 60^{circ} to horizontal. kinetic energy o

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3-2.Motion in Plane

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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.

A$Zero$

B$\frac{E}{2}$

C$\frac{E}{4}$

D$E$

(JEE MAIN-2022)

Solution

$E =\frac{1}{2} mu ^{2}$
At Highest point, Velocity $V = u \cos 60^{\circ}=\frac{ u }{2}$
$\therefore K E \text { at topmost point }=\frac{1}{2} m \left(\frac{ u }{2}\right)^{2}=\frac{E}{4}$

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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.

Solution

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