A ball having kinetic energy KE is projected | vedclass

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(B) The initial kinetic energy is given by $KE = \frac{1}{2} m u^2$,where $u$ is the initial velocity.At the highest point of the trajectory,the verti

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$\frac{KE}{4}$

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(B) The initial kinetic energy is given by $KE = \frac{1}{2} m u^2$,where $u$ is the initial velocity.At the highest point of the trajectory,the vertical component of velocity becomes zero,and the velocity of the ball is equal to its horizontal component.$v_x = u \cos 	heta = u \cos 60^{\circ} = u 	imes \frac{1}{2} = \frac{u}{2}$.The kinetic energy at the highest point $(KE_{top})$ is given by:$KE_{top} = \frac{1}{2} m v_x^2 = \frac{1}{2} m \left( \frac{u}{2} ight)^2$.$KE_{top} = \frac{1}{2} m \left( \frac{u^2}{4} ight) = \frac{1}{4} \left( \frac{1}{2} m u^2 ight)$.Since $KE = \frac{1}{2} m u^2$,we have $KE_{top} = \frac{KE}{4}$.

Column-$I$ $(R/H_{max})$	Column-$II$ (Angle of projection $\theta$)
$A. 1$	$1. 60^o$
$B. 4$	$2. 30^o$
$C. 4\sqrt{3}$	$3. 45^o$
$D. 4/\sqrt{3}$	$4. \tan^{-1}(4) = 76^o$

$A$ ball having kinetic energy $KE$ is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of the ball at the highest point of its flight?

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$A. 1$ $1. 60^o$

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