A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

Motion in A Plane PYQ — JEE Main 2004 · Shift 0 · Q186 with Solution

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Correct answer: A

For a projectile projected with speed $u$ at angle $\theta$ , the horizontal range is

R = \frac{u^2 \sin 2\theta}{g}.

For the same range with the same speed, there are two possible angles:

\theta \quad \text{and} \quad 90^\circ-\theta,

because

\sin 2\theta = \sin\big(2(90^\circ-\theta)\big)=\sin(180^\circ-2\theta)=\sin 2\theta.

Time of flight for angle $\theta$ is

T_1 = \frac{2u\sin\theta}{g}.

For angle $90^\circ-\theta$ ,

T_2 = \frac{2u\sin(90^\circ-\theta)}{g} = \frac{2u\cos\theta}{g}.

Multiply:

T_1T_2 = \left(\frac{2u\sin\theta}{g}\right)\left(\frac{2u\cos\theta}{g}\right) = \frac{4u^2\sin\theta\cos\theta}{g^2}.

Using

2\sin\theta\cos\theta = \sin 2\theta,

we get

T_1T_2 = \frac{2u^2\sin 2\theta}{g^2}.

Now from the range formula,

R = \frac{u^2\sin 2\theta}{g} \implies u^2\sin 2\theta = Rg.

Substitute into the product:

T_1T_2 = \frac{2Rg}{g^2} = \frac{2R}{g}.

Since

T_1T_2 = \frac{2R}{g},

the product $T_1T_2$ is directly proportional to $R$ .

Therefore, the correct option is:

\boxed{A}\text{ : }R

Stored correct answer: $A$

Our derived answer: $A$

They match.