Question

If two stones are projected at angle \( \theta \) and \( (90^\circ - \theta) \) respectively with horizontal with a speed of \( 20 \, {m/s} \). If the second stone rises \( 10 \, {m} \) higher than the first stone, then the angle of projection \( \theta \) is (acceleration due to gravity = \( 10 \, {m/s}^2 \)):

• A. \( 45^\circ \)
• B. \( 30^\circ \)
• C. \( 60^\circ \)
• D. \( 20^\circ \)

Hint:

For projectile motion problems involving angles, recall that complementary angles yield the same range, but different heights.

Answer 1

The Correct Option is B

Given two projectiles launched at complementary angles \( \theta \) and \( 90^\circ - \theta \) with the same initial velocity \( 20 \, {m/s} \). 
The range of both projectiles would be the same due to the property of complementary angles having the same range. 
However, their maximum heights will differ. 
The height \( h \) for a projectile launched at angle \( \theta \) with initial velocity \( v \) is given by: \[ h = \frac{v^2 \sin^2(\theta)}{2g} \] Where \( g = 10 \, {m/s}^2 \). For \( \theta \) and \( 90^\circ - \theta \): \[ h_1 = \frac{400 \sin^2(\theta)}{20} \] \[ h_2 = \frac{400 \cos^2(\theta)}{20} \] Given \( h_2 = h_1 + 10 \): \[ \frac{400 \cos^2(\theta)}{20} = \frac{400 \sin^2(\theta)}{20} + 10 \] \[ 20 \cos^2(\theta) = 20 \sin^2(\theta) + 10 \] \[ 20 (1 - \sin^2(\theta)) = 20 \sin^2(\theta) + 10 \] \[ 20 = 40 \sin^2(\theta) + 10 \] \[ 10 = 40 \sin^2(\theta) \] \[ \sin^2(\theta) = \frac{1}{4} \] \[ \sin(\theta) = \frac{1}{2} \] Thus, \( \theta = 30^\circ \).