If T1 and T2 be the times of flight for a projectile havng same range R for two angles of projection The product (T1)*(T2) is proportional to _

R , R^@ , 1/R , 1/R^2

To determine the relationship between the times of flight \( T_1 \) and \( T_2 \) for a projectile launched at two different angles but achieving the same range \( R \), we can delve into the physics of projectile motion. The key here is to understand how the time of flight and range are related through the equations of motion.

Understanding Projectile Motion

When a projectile is launched, its motion can be analyzed using the equations of kinematics. The range \( R \) of a projectile launched at an angle \( \theta \) with an initial velocity \( v_0 \) is given by the formula:

• R = (v_0^2 * sin(2θ)) / g

Here, \( g \) is the acceleration due to gravity. The time of flight \( T \) for a projectile is given by:

• T = (2 * v_0 * sin(θ)) / g

Analyzing Two Angles

Let’s denote the two angles of projection as \( \theta_1 \) and \( \theta_2 \). Since both angles yield the same range \( R \), we can express the relationship between the two angles using the range formula:

• R = (v_0^2 * sin(2θ_1)) / g = (v_0^2 * sin(2θ_2)) / g

This implies that:

• sin(2θ_1) = sin(2θ_2)

From trigonometric identities, we know that \( \theta_1 \) and \( \theta_2 \) can be complementary angles (i.e., \( \theta_1 + \theta_2 = 90^\circ \)).

Calculating the Product of Times of Flight

Now, let’s calculate the product of the times of flight:

• T1 = (2 * v_0 * sin(θ_1)) / g
• T2 = (2 * v_0 * sin(θ_2)) / g

The product \( T_1 \cdot T_2 \) can be expressed as:

• T1 * T2 = [(2 * v_0 * sin(θ_1)) / g] * [(2 * v_0 * sin(θ_2)) / g]
• = (4 * v_0^2 * sin(θ_1) * sin(θ_2)) / g^2

Using the identity for sine of complementary angles, we know that:

• sin(θ_2) = cos(θ_1)

Thus, we can rewrite the product:

• T1 * T2 = (4 * v_0^2 * sin(θ_1) * cos(θ_1)) / g^2

Using the double angle identity, we have:

• sin(2θ_1) = 2 * sin(θ_1) * cos(θ_1)

Substituting this back, we get:

• T1 * T2 = (2 * v_0^2 * sin(2θ_1)) / g^2

Relating to Range

Now, since we know that \( R \) is proportional to \( v_0^2 \) and \( sin(2θ_1) \), we can express \( v_0^2 \) in terms of \( R \):

• R = (v_0^2 * sin(2θ_1)) / g
• => v_0^2 = (R * g) / sin(2θ_1)

Substituting this back into our equation for \( T_1 \cdot T_2 \):

• T1 * T2 = (2 * (R * g) / sin(2θ_1) * sin(2θ_1)) / g^2
• = (2R) / g

This shows that the product of the times of flight \( T_1 \cdot T_2 \) is directly proportional to the range \( R \).

Final Result

In conclusion, the product \( T_1 \cdot T_2 \) is proportional to \( R \). Therefore, the correct answer to your question is:

• R