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.

• A. ${\theta }_1+{\theta }_2={90}^{\circ }$
• B. $\frac{t_1}{t_2}=\tan {\theta }_1$
• C. $\frac{t_1}{t_2}=\tan {\theta }_2$
• D. $\frac{t_1}{\sin {\theta }_1}=\frac{t_2}{\sin {\theta }_2}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A ${\theta }_1+{\theta }_2={90}^{\circ }$
B $\frac{t_1}{t_2}=\tan {\theta }_1$
D $\frac{t_1}{\sin {\theta }_1}=\frac{t_2}{\sin {\theta }_2}$

As both projectiles have same speed and horizontal range so it is complementary angles projectile motion. For this motion, ${\theta }_1+{\theta }_2={90}^o$
Time for flight $t=\frac{2u\sin \theta }{g}$
so, $t_1=\frac{2u\sin {\theta }_1}{g}$ and $t_2=\frac{2u\sin {\theta }_2}{g}$
thus, $\frac{t_1}{t_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin {\theta }_1}{\sin (90-{\theta }_1)}=\frac{\sin {\theta }_1}{\cos {\theta }_1}=\tan {\theta }_1$