Question

A body is projected at an angle of \(60^\circ\) with the horizontal. If the initial kinetic energy of the body is \(X\), then its kinetic energy at the highest point is

• A. \(X\)
• B. \(2X\)
• C. \(\frac{X}{2}\)
• D. \(\frac{X}{4}\)

Hint:

At the peak of a projectile’s path, vertical velocity is zero. So, use only the horizontal component to calculate kinetic energy.

Answer 1

The Correct Option is D

Step 1: Understand the kinetic energy at the highest point 
At the highest point of projectile motion, the vertical component of velocity becomes zero. Only the horizontal component of velocity contributes to the kinetic energy. 
Step 2: Resolve initial velocity 
Let the initial velocity be \(u\). The initial kinetic energy is: \[ X = \frac{1}{2} m u^2 \] The horizontal component of velocity is: \[ u_x = u \cos 60^\circ = \frac{u}{2} \] Step 3: Kinetic energy at the highest point 
At the highest point, only horizontal velocity exists: \[ KE = \frac{1}{2} m u_x^2 = \frac{1}{2} m \left( \frac{u}{2} \right)^2 = \frac{1}{2} m \cdot \frac{u^2}{4} = \frac{1}{4} \cdot \frac{1}{2} m u^2 = \frac{X}{4} \]