# It is possible to project a particle with a given speed in two possible ways so that it has the same horizontal range 'R'. The product of time taken by it in the two possible ways is: A. R/g B. 2R/g C. 3R/g D. 4R/g

## Question

It is possible to project a particle with a given speed in two possible ways so that it has the same horizontal range 'R'. The product of time taken by it in the two possible ways is:

A. R/g B. 2R/g C. 3R/g D. 4R/g

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## Answer

To solve the problem, let's first understand the variables and equations involved:

Given:

- A particle can be projected with a given speed in two possible ways to achieve the same horizontal range ( R ).
- We need to find the product of the times taken for the two different projections.

To start, we know the formula for the horizontal range of a projectile: $$ R = \frac{v_0^2 \sin 2\theta}{g} $$ where ( v_0 ) is the initial velocity, ( \theta ) is the angle of projection, and ( g ) is the acceleration due to gravity.

Since the horizontal range ( R ) remains the same for both projections, we have: $$ R_1 = R_2 $$

For two different angles ( \theta_1 ) and ( \theta_2 ) that result in the same horizontal range, the relationship between these angles can be given by: $$ \sin 2\theta_1 = \sin 2\theta_2 $$

This implies that: $$ 2\theta_1 = 180^\circ - 2\theta_2 $$ or $$ \theta_1 = 90^\circ - \theta_2 $$

Next, we calculate the time of flight for each projection. The time of flight ( T ) is given by: $$ T = \frac{2 v_0 \sin\theta}{g} $$

For two angles ( \theta_1 ) and ( \theta_2 ), the times of flight are: $$ T_1 = \frac{2 v_0 \sin\theta_1}{g} $$ $$ T_2 = \frac{2 v_0 \sin\theta_2}{g} $$

To find the product ( T_1 \times T_2 ), multiply the two times: $$ T_1 \times T_2 = \left( \frac{2 v_0 \sin\theta_1}{g} \right) \times \left( \frac{2 v_0 \sin\theta_2}{g} \right) $$ $$ = \frac{4 v_0^2 \sin\theta_1 \sin\theta_2}{g^2} $$

Using the fact that ( \theta_1 = 90^\circ - \theta_2 ), we can substitute for ( \sin(90^\circ - \theta) = \cos\theta ): $$ T_1 \times T_2 = \frac{4 v_0^2 \sin\theta_1 \cos\theta_1}{g^2} $$ $$ = \frac{4 v_0^2 \frac{\sin 2\theta_1}{2}}{g^2} $$ $$ = \frac{2 v_0^2 \sin 2\theta_1}{g^2} $$

Given ( v_0^2 \sin 2\theta_1 = Rg ): $$ T_1 \times T_2 = \frac{2Rg}{g^2} $$ $$ = \frac{2R}{g} $$

Thus, the product of the time taken for the two possible ways is: $$ \boxed{\frac{2R}{g}}

The correct answer is **B. ( \frac{2R}{g} )**.

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