(A) The initial kinetic energy of the body is $K = \frac{1}{2} m u^2$.At the highest point of the trajectory,the vertical component of velocity is zer

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(A) The initial kinetic energy of the body is $K = \frac{1}{2} m u^2$.At the highest point of the trajectory,the vertical component of velocity is zero,and the horizontal component remains $u \cos \beta$.Thus,the kinetic energy at the highest point is $K' = \frac{1}{2} m (u \cos \beta)^2 = K \cos^2 \beta$.Given that $K' = \frac{3}{4} K$,we have $K \cos^2 \beta = \frac{3}{4} K$.This simplifies to $\cos^2 \beta = \frac{3}{4}$,which means $\cos \beta = \frac{\sqrt{3}}{2}$.Therefore,$\beta = 30^{\circ}$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

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$A$ body is projected with a speed $u \ m/s$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3/4$ of the initial kinetic energy. The value of $\beta$ is: (in $^{\circ}$)

WBJEE 2010 All WBJEE

A
$30$
B
$45$
C
$60$
D
$120$

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$A$ body is projected with a speed $u \ m/s$ at an angle $\beta$ with the horizontal. The kinetic energy at the highest point is $3/4$ of the initial kinetic energy. The value of $\beta$ is: (in $^{\circ}$)

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