Question

If two bodies A and B are projected with same velocity but with different angles \( \theta_1 \) and \( \theta_2 \) respectively with the horizontal such that both will have same range, then the ratio of times of flight of the bodies A and B is

• A. \( \sin \theta_2 \)
• B. \( \sin \theta_1 \)
• C. \( \tan \theta_2 \)
• D. \( \tan \theta_1 \)

Hint:

When two projectiles are launched with same speed and same range, their angles of projection are complementary. Use this to relate sine and cosine in time of flight ratio.

Answer 1

The Correct Option is D

Step 1: Recall the formula for range of a projectile. \[ R = \frac{u^2 \sin(2\theta)}{g} \] If two projectiles have the same initial speed and same range, and are projected at angles \( \theta_1 \) and \( \theta_2 \), then: \[ \sin(2\theta_1) = \sin(2\theta_2) \] This implies: \[ 2\theta_1 + 2\theta_2 = 180^\circ \Rightarrow \theta_1 + \theta_2 = 90^\circ \] Step 2: Recall the time of flight formula. \[ T = \frac{2u \sin\theta}{g} \] Let \( T_A \) and \( T_B \) be times of flight of bodies A and B: \[ \frac{T_A}{T_B} = \frac{\sin\theta_1}{\sin\theta_2} \] Since \( \theta_1 + \theta_2 = 90^\circ \Rightarrow \theta_2 = 90^\circ - \theta_1 \), so: \[ \sin \theta_2 = \cos \theta_1 \] \[ \frac{T_A}{T_B} = \frac{\sin \theta_1}{\cos \theta_1} = \tan \theta_1 \] % Final Answer \[ \boxed{\tan \theta_1} \]