Question

Two particles are projected from the same point with the same speed, at different angles $\theta_{1}$ and $\theta_{2}$ to the horizontal. They have the same horizontal range. Their times of flights are $t_{1}$ and $t_{2}$ respectively. Then
(1) $\theta_{1}+\theta_{2}=90^{\circ}$
(2) $\frac{t_{1}}{\sin \theta_{1}}=\frac{t_{2}}{\sin \theta_{2}}$
(3) $\frac{t_{1}}{t_{2}}=\tan \theta_{1}$
(4) $\frac{t_{1}}{t_{2}}=\tan \theta_{2}$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (A)

Projectiles have the same horizontal range for complimentary angles of projection.
$\therefore \theta_{1}+\theta_{2}=90^{\circ} $
$ \theta_{2}=90^{\circ}-\theta_{1}$
The time of flight for first particle;
$t_{1}=\frac{2 u \sin \theta_{1}}{g}$
The time of flight for second particle
$t_{2}=\frac{2 u \sin \theta_{2}}{g} $
$\therefore \frac{t_{1}}{t_{2}}=\frac{\sin \theta_{1}}{\sin \theta_{2}} t_{1}$
Or $\frac{t_{1}}{\sin \theta_{1}}=\frac{t_{2}}{\sin \theta_{2}}$
Again, $\frac{t_{1}}{t_{2}}=\frac{\sin \theta_{1}}{\sin \theta_{2}}$
Or $\frac{t_{1}}{t_{2}}=\frac{\sin \theta_{1}}{\sin \left(90-\theta_{1}\right)}$
$\frac{t_{1}}{t_{2}}=\frac{\sin \theta_{1}}{\cos \left(\theta_{1}\right)}$
$\therefore \frac{t_{1}}{t_{2}}=\tan \theta_{1}$