Question

A ball is projected at an angle of ${30}^{\circ }$ with horizontal with kinetic energy E. The kinetic energy at the highest point is $k\times E$. The Value of k (correct upto two decimal places) is ___

• A. 0.75

Answer 1

The correct option is A 0.75

We know
$H=\frac{u_0^2{\sin }^2{30}^{\circ }}{2g}$
At point A, $kE=T.E=\frac{1}{2}m{u}_0^2$
At point B, $K.E=\frac{1}{2}\times m\times (u_0\cos {30}^{\circ }{)}^2=\frac{3}{4}\times \frac{1}{2}m{u}_0^2$
$=0.75\times E$
$=K\times E$
$\Rightarrow K=0.75$