Question

A projectile can have the same range \( R \) for two angles of projection. Their initial velocities are the same. If \( T_1 \) and \( T_2 \) are times of flight in two cases, then the product of two times of flight is directly proportional to:

• A. \( \frac{1}{R} \)
• B. \( R^3 \)
• C. \( R^2 \)
• D. \( R \)

Hint:

N/A

Answer 1

The Correct Option is D

We are given that a projectile can have the same range \( R \) for two angles of projection with the same initial velocity. Let \( T_1 \) and \( T_2 \) be the times of flight for two different angles of projection. We are asked to find the relationship between the product of the two times of flight and the range \( R \). The time of flight \( T \) for a projectile is given by the formula: \[ T = \frac{2u \sin \theta}{g}, \] where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. The range \( R \) of the projectile is given by: \[ R = \frac{u^2 \sin 2\theta}{g}. \] Since the initial velocity \( u \) is the same for both cases, we can write: \[ R = \frac{u^2 \sin 2\theta}{g} \quad \text{and} \quad T = \frac{2u \sin \theta}{g}. \] The product of the times of flight for two angles is: \[ T_1 \times T_2 = \left(\frac{2u \sin \theta_1}{g}\right) \left(\frac{2u \sin \theta_2}{g}\right). \] We can simplify this expression as: \[ T_1 \times T_2 = \frac{4u^2 \sin \theta_1 \sin \theta_2}{g^2}. \] Using the identity \( \sin \theta_1 \sin \theta_2 = \frac{1}{2} [\cos(\theta_1 - \theta_2) - \cos(\theta_1 + \theta_2)] \) and knowing that the range \( R \) is directly proportional to \( \sin 2\theta \), we can conclude that the product of the times of flight is directly proportional to \( R \). Thus, the correct answer is option (4).

Answer 2

Step 1: General range formula 

For a projectile launched with velocity \( u \) at angle \( \theta \): \[ R = \frac{u^2 \sin 2\theta}{g} \]

Step 2: Time of flight formula

Time of flight for angle \( \theta \): \[ T = \frac{2u \sin \theta}{g} \] Let the two angles be \( \theta \) and \( 90^\circ - \theta \). They will have the same range because: \[ \sin 2\theta = \sin \left[2(90^\circ - \theta)\right] = \sin (180^\circ - 2\theta) = \sin 2\theta \]

Step 3: Compute product of the two times of flight

\[ T_1 = \frac{2u \sin \theta}{g}, \quad T_2 = \frac{2u \cos \theta}{g} \] So the product: \[ T_1 T_2 = \left(\frac{2u \sin \theta}{g}\right) \cdot \left(\frac{2u \cos \theta}{g}\right) = \frac{4u^2 \sin \theta \cos \theta}{g^2} \] Recall: \[ \sin 2\theta = 2 \sin \theta \cos \theta \Rightarrow \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \] Thus, \[ T_1 T_2 = \frac{4u^2}{g^2} \cdot \frac{1}{2} \sin 2\theta = \frac{2u^2 \sin 2\theta}{g^2} \]

Step 4: Express in terms of range \( R \)

Since: \[ R = \frac{u^2 \sin 2\theta}{g} \Rightarrow \frac{2u^2 \sin 2\theta}{g^2} = \frac{2R}{g} \] So, \[ T_1 T_2 \propto R \]

Answer:

\( \boxed{R} \)