Question

A projectile with speed \( 50\, \text{ms}^{-1} \) is thrown at an angle of \( 60^\circ \) with the horizontal. The maximum height that can be reached is (take \( g = 10\, \text{ms}^{-2} \))

• A. 90.75 m
• B. 70.00 m
• C. 85.00 m
• D. 93.75 m

Hint:

Use \( H = \frac{u^2 \sin^2 \theta}{2g} \) for vertical displacement in projectile motion.

Answer 1

The Correct Option is D

Maximum height in projectile motion is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] Here, \( u = 50 \), \( \theta = 60^\circ \), \( g = 10 \).
\[ H = \frac{50^2 \cdot \sin^2 60^\circ}{2 \cdot 10} = \frac{2500 \cdot \left( \frac{3}{4} \right)}{20} = \frac{1875}{20} = 93.75 \, \text{m} \]