Question

Two particles are projected at angles \( \theta \) and \( (90^\circ - \theta) \) with the same speed of 25 m/s. If the second particle reaches 15 m higher than the first, then the angle of projection \( \theta \) is:

• A. \( 15^\circ \)
• B. \( 30^\circ \)
• C. \( 45^\circ \)
• D. \( 60^\circ \)

Hint:

In such projectile problems, always compare the expressions for maximum height and apply trigonometric identities to simplify.

Answer 1

The Correct Option is B

Step 1: Use the formula for maximum height \[ H = \frac{u^2 \sin^2\theta}{2g} \] For the first particle: \[ H_1 = \frac{u^2 \sin^2 \theta}{2g} \] For the second particle (projected at \( 90^\circ - \theta \)): \[ H_2 = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g} \] Step 2: Use the given height difference \[ H_2 - H_1 = 15 \Rightarrow \frac{u^2}{2g}(\cos^2 \theta - \sin^2 \theta) = 15 \] Use the identity: \[ \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \] \[ \Rightarrow \frac{u^2}{2g} \cos(2\theta) = 15 \] Substitute \( u = 25 \ \text{m/s}, \ g = 10 \ \text{m/s}^2 \): \[ \frac{625}{20} \cos(2\theta) = 15 \Rightarrow 31.25 \cos(2\theta) = 15 \Rightarrow \cos(2\theta) = \frac{15}{31.25} = 0.48 \] Step 3: Solve for \( \theta \) \[ 2\theta = \cos^{-1}(0.48) \approx 61.4^\circ \Rightarrow \theta \approx \frac{61.4^\circ}{2} \approx 30.7^\circ \] \[ \boxed{\theta \approx 30^\circ} \]