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  4. 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

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Q. A ball is projected with kinetic energy $E$ at an angle of $45^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

Solution:

$E^{'}=E \cos ^{2} \theta=E \cos ^{2}\left(45^{\circ}\right)=\frac{E}{2}$