Question

A particle is projected from ground with speed u and at an angle \(\theta\) with horizontal. If at maximum height from ground, the speed of a particle is half times its initial velocity of projection then its maximum height attained is \((\cos 60^{\circ }=0.5,\,\sin 60^{\circ }=\sqrt{3}/2)\) (g = acceleration due to gravity)

• A. \(\frac{2u^{2}}{g}\)
• B. \(\frac{3u^{2}}{8g}\)
• C. \(\frac{u^{2}}{g}\)
• D. \(\frac{u^{2}}{2g}\)

Hint:

At the top only the horizontal component remains, which fixes the angle of projection.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At maximum height the vertical velocity is zero, so the speed there equals the horizontal component \(u\cos\theta\).

Step 2: Find the Angle
Given \(u\cos\theta=\frac{u}{2}\), so \(\cos\theta=0.5\) and \(\theta=60^{\circ}\).

Step 3: Use the Height Formula
\(H=\frac{u^{2}\sin^{2}\theta}{2g}\)

Step 4: Calculation
With \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}\), \(H=\frac{u^{2}\times\frac{3}{4}}{2g}=\frac{3u^{2}}{8g}\).

Final Answer:
The maximum height reached is 3u squared over 8g. \[ \boxed{\text{(B) }\dfrac{3u^{2}}{8g}} \]