A stone is projected at an angle of 30^ to the horizontal. The ratio of kinetic energy at projection to its kinetic ener

Question

A stone is projected at an angle of 30^ to the horizontal. The ratio of kinetic energy at projection to its kinetic energy at the highest point is:

Accepted Answer

Given: A stone is projected with initial speed $u$ at an angle of $30^\circ$ .

Find: The ratio of kinetic energy at projection to the kinetic energy at the highest point.

At projection, the kinetic energy is

KE_{\text{projection}} = \frac{1}{2}mu^2

At the highest point of projectile motion, the vertical component of velocity becomes zero, so only the horizontal component remains. Therefore,

KE_{\text{highest}} = \frac{1}{2}m\left(u\cos 30^\circ\right)^2

Now take the ratio:

\frac{KE_{\text{projection}}}{KE_{\text{highest}}} = \frac{\frac{1}{2}mu^2}{\frac{1}{2}m\left(u\cos 30^\circ\right)^2} = \frac{u^2}{u^2\cos^2 30^\circ} = \frac{1}{\cos^2 30^\circ}

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ ,

\frac{1}{\cos^2 30^\circ} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3}

Therefore, the required ratio is $4:3$ . The solution states the correct option is A, but the worked value matches option D.

Answer