Question

A particle is projected at an angle $30^\circ$ with horizontal having kinetic energy $K$. The kinetic energy of the particle at the highest point is.

• A. \( \frac{1}{2} K \)
• B. \( \frac{3}{4} K \)
• C. \( \frac{3}{8} K \)
• D. \( \frac{5}{8} K \)

Hint:

At the highest point of projectile motion, only the horizontal component of velocity contributes to the kinetic energy. The vertical component becomes zero at this point.

Answer 1

The Correct Option is B

To determine the kinetic energy at the highest point of the projectile's motion, we need to consider both the initial kinetic energy and the nature of projectile motion. When a particle is projected at an angle $\theta$ with initial kinetic energy $K$, the initial velocity components are:

• Horizontal component: $v_{0x} = v_0 \cos \theta$
• Vertical component: $v_{0y} = v_0 \sin \theta$

At the highest point of its trajectory, the vertical component of velocity becomes zero, while the horizontal component remains unchanged due to the absence of horizontal forces (assuming air resistance is negligible). 

Therefore, the kinetic energy at the highest point depends only on this horizontal component. Initial kinetic energy, \( K \), is given by:

\( K = \frac{1}{2} m v_0^2 \)

The horizontal kinetic energy at the highest point is:

\( KE_{\text{horizontal}} = \frac{1}{2} m (v_0 \cos \theta)^2 \)

Substituting \( v_0^2 = \frac{2K}{m} \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), we have:

\( KE_{\text{horizontal}} = \frac{1}{2} m \left( \frac{\sqrt{3}}{2}v_0 \right)^2 = \frac{1}{2} m \cdot \frac{3}{4} v_0^2 \)

\( KE_{\text{horizontal}} = \frac{3}{8} m v_0^2 \)

Substitute \( v_0^2 = \frac{2K}{m} \):

\( KE_{\text{horizontal}} = \frac{3}{8} \cdot \frac{2K}{m} \cdot m = \frac{3}{4} K \)

This shows that the kinetic energy at the highest point is \(\frac{3}{4} K\).

Answer 2

When a particle is projected at an angle, its total kinetic energy at any point is the sum of its horizontal and vertical kinetic energies. At the highest point of the trajectory, the vertical velocity becomes zero, and only the horizontal velocity remains. The horizontal velocity \( v_x \) is given by: \[ v_x = v_0 \cos(\theta) \] where \( \theta = 30^\circ \) is the angle of projection and \( v_0 \) is the initial velocity. The total initial kinetic energy is: \[ K = \frac{1}{2} m v_0^2 \] At the highest point, the kinetic energy is due only to the horizontal component of velocity: \[ K_{\text{highest}} = \frac{1}{2} m v_x^2 = \frac{1}{2} m (v_0 \cos(30^\circ))^2 = \frac{1}{2} m v_0^2 \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} K \]
Thus, the kinetic energy at the highest point is \( \frac{3}{4} K \).