JEE PYQ: Motion in a Plane - Question ID f181bb916654 (JEE Main 2019)

In projectile motion on level ground, a shell fired with initial speed $u$ at an angle $\theta$ to the horizontal travels a horizontal range $R = \frac{u^2 \sin 2\theta}{g}.$ The time of flight for each trajectory is $t = \frac{2u \sin\theta}{g}.$ When the gun is fixed and the target is at range $R$, there are exactly two launch angles $\theta$ and $(90^\circ - \theta)$ that give the same range. These two angles produce two different times of flight $t_1$ and $t_2$.

1. Write the two expressions for the times of flight: $t_1 = \frac{2u \sin\theta}{g}, \quad t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos\theta}{g}.$ 2. Form the product: $t_1 t_2 = \frac{2u \sin\theta}{g} \cdot \frac{2u \cos\theta}{g} = \frac{4u^2 \sin\theta \cos\theta}{g^2} = \frac{2u^2 \sin 2\theta}{g^2}.$ 3. Recall the range formula: $R = \frac{u^2 \sin 2\theta}{g}.$ Hence $\frac{2u^2 \sin 2\theta}{g^2} = \frac{2R}{g}.$ 4. Therefore $t_1 t_2 = \frac{2R}{g}.$

• Many students try to solve for $\theta$ explicitly, which is unnecessary and time-consuming. • Forgetting that there are two angles (complementary) giving the same range can lead to missing one of the times. • A sign error in the trigonometric identity $\sin 2\theta = 2\sin\theta\cos\theta$ can spoil the product. Always remember: the product of the two times of flight for the same range is independent of the launch angle and depends only on $R$ and $g$.