Question

A particle is projected with a velocity of 20 m/s at an angle of 30° to the horizontal. What is the maximum height reached? (Take \( g = 10 \, \text{m/s}^2 \)).

• A. 5 m
• B. 10 m
• C. 15 m
• D. 20 m

Hint:

For projectile motion, use \( H = \frac{u^2 \sin^2 \theta}{2g} \) to find the maximum height, ensuring the angle is in degrees and \( \sin \theta \) is calculated correctly.

Answer 1

The Correct Option is A

The maximum height of a projectile is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g} \] where \( u = 20 \, \text{m/s} \), \( \theta = 30^\circ \), and \( g = 10 \, \text{m/s}^2 \). Since \( \sin 30^\circ = \frac{1}{2} \), we have: \[ \sin^2 30^\circ = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \] \[ H = \frac{20^2 \cdot \frac{1}{4}}{2 \cdot 10} = \frac{400 \cdot \frac{1}{4}}{20} = \frac{100}{20} = 5 \, \text{m} \] Thus, the maximum height is: \[ \boxed{5} \]