Question

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

• A. 4 : 3
• B. 4 : 1
• C. 1 : 2
• D. 1 : 4

Hint:

For projectile motion:
• At the highest point, only horizontal velocity contributes to kinetic energy.
• Use trigonometric components to relate initial and final kinetic energies.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the kinetic energy of a projectile at its point of projection and at its highest point.

When a stone is projected at an angle \(\theta = 30\degree\) with an initial velocity \(u\), the kinetic energy at the point of projection (\(KE_1\)) is given by:

KE_1 = \frac{1}{2} m u^2

At the highest point of the projectile's path, the vertical component of the velocity is zero, and only the horizontal component contributes to the kinetic energy. The horizontal component of velocity (\(u_x\)) is:

u_x = u \cos \theta

Therefore, the kinetic energy at the highest point (\(KE_2\)) is:

KE_2 = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m (u^2 \cos^2 \theta)

Now, we can find the ratio of kinetic energy at the point of projection to the kinetic energy at the highest point:

\text{Ratio} = \frac{KE_1}{KE_2} = \frac{\frac{1}{2} m u^2}{\frac{1}{2} m u^2 \cos^2 \theta}

After simplifying, we get:

\text{Ratio} = \frac{1}{\cos^2 \theta}

Substituting \(\theta = 30\degree\), we know:

\cos 30\degree = \frac{\sqrt{3}}{2}

Therefore:

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

Thus, the ratio becomes:

\text{Ratio} = \frac{1}{\frac{3}{4}} = \frac{4}{3}

Therefore, the ratio of kinetic energy at projection to its kinetic energy at the highest point is \(4 : 3\).

The correct answer is: 4 : 3.