Question

A particle is projected at \(60^{\circ }\) to the horizontal with kinetic energy 'K'. The kinetic energy at the highest point is \(\left(\cos 60^{0}=\frac{1}{2}\right)\)

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

Hint:

At the top only the horizontal velocity component remains.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
At the highest point the vertical velocity is zero. Only the horizontal component \(v\cos\theta\) remains.

Step 2: Calculation
Initial KE: \(K=\dfrac{1}{2}mv^{2}\).
KE at top \(=\dfrac{1}{2}m(v\cos60^{\circ})^{2}=K\cos^{2}60^{\circ}\).

Step 3: Substitute
\(\cos60^{\circ}=\dfrac{1}{2}\), so KE at top \(=K\times\dfrac{1}{4}=\dfrac{K}{4}\).

Final Answer:
The kinetic energy at the highest point is K/4. \[ \boxed{\text{(D) }\dfrac{K}{4}} \]