Question

A particle is projected making an angle of 45$^{\circ}$ with horizontal having kinetic energy K. The kinetic energy at highest point will be

• A. $\frac{K}{\sqrt 2}$
• B. $\frac{K}{2}$
• C. 2K
• D. K

Answer 1

The Correct Option is B

To solve this problem, we must consider the projectile motion of the particle and how kinetic energy is distributed during its flight.

Initially, when the particle is projected, it has a total kinetic energy, K, that is distributed between its horizontal and vertical components. At the highest point of its trajectory, the vertical component of velocity becomes zero, and all kinetic energy is due to horizontal motion.

The velocity of the particle can be split into two components at the point of projection:

• Horizontal component: v_x = v \cos \theta
• Vertical component: v_y = v \sin \theta

Given that the angle of projection \theta is 45^\circ, we have:

• v_x = v \cos 45^\circ = \frac{v}{\sqrt 2}
• v_y = v \sin 45^\circ = \frac{v}{\sqrt 2}

Using the definition of kinetic energy, we have:

• Initial kinetic energy: K = \frac{1}{2} m v^2

At the highest point, the vertical velocity component becomes zero, and only the horizontal component contributes to the kinetic energy:

• Horizontal kinetic energy at the highest point: K_{\text{highest}} = \frac{1}{2} m (v_x)^2 = \frac{1}{2} m \left( \frac{v}{\sqrt{2}} \right)^2 = \frac{1}{2} m \cdot \frac{v^2}{2} = \frac{1}{2} \left( \frac{1}{2} m v^2 \right)

Simplifying, we find:

• K_{\text{highest}} = \frac{K}{2}

Therefore, the kinetic energy at the highest point of the particle's projectile motion is \frac{K}{2}.

Thus, the correct answer is \frac{K}{2}.