A body is projected with a speed u m/s at an | vedclass

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(A) The initial kinetic energy at the point of projection $O$ is given by:$K = \frac{1}{2} mu^2$where $m$ is the mass of the body and $u$ is the initi

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$30$

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(A) The initial kinetic energy at the point of projection $O$ is given by:$K = \frac{1}{2} mu^2$where $m$ is the mass of the body and $u$ is the initial velocity of projection.At the highest point (i.e.,at maximum height $H$),the vertical component of velocity becomes zero,and the velocity is only the horizontal component:$v = u \cos \beta$Therefore,the kinetic energy at the highest point is:$K^{\prime} = \frac{1}{2} mv^2 = \frac{1}{2} m(u \cos \beta)^2 = \frac{1}{2} mu^2 \cos^2 \beta$According to the problem,the kinetic energy at the highest point is $\frac{3}{4}$ of the initial kinetic energy:$K^{\prime} = \frac{3}{4} K$Substituting the expressions:$\frac{1}{2} mu^2 \cos^2 \beta = \frac{3}{4} \left( \frac{1}{2} mu^2 \right)$$\cos^2 \beta = \frac{3}{4}$Taking the square root on both sides:$\cos \beta = \frac{\sqrt{3}}{2}$$\beta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = 30^{\circ}$

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

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