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;
$t_1=\frac{2u\sin {\theta }_1}{g}$
The time of flight for second particle
$t_2=\frac{2u\sin {\theta }_2}{g}$
$\therefore \frac{t_1}{t_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}{t}_1$
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 $\frac{t_1}{t_2}=\frac{\sin {\theta }_1}{\sin \left(90-{\theta }_1\right)}$
$\frac{t_1}{t_2}=\frac{\sin {\theta }_1}{\cos \left({\theta }_1\right)}$
$\therefore \frac{t_1}{t_2}=\tan {\theta }_1$