Question

Trajectories of two projectiles are shown in figure. Let $T_1$ and $T_2$ be the time periods and $u_1$ and $u_2$ their speeds of projection. Then

• A. $T_1=T_2$
• B. $T_1>T_2$
• C. $T_1<T_2$
• D. Can not say

Answer 1

The correct option is A $T_1=T_2$

From the diagram we can see, the height of both the projectiles are same. hence $H_1=H_2$
as we know, $H=\frac{u^2{\sin }^2\theta }{2g}$
(where H is maximum height of projectile, $u$ is initial speed and $\theta$ is the angle of projectile.
so, $\frac{u_1^2{\sin }^2{\theta }_1}{2g}=\frac{u_2^2{\sin }^2{\theta }_2}{2g}$
$u_1^2{\sin }^2{\theta }_1=u_2^2{\sin }^2{\theta }_2$
$u_1\sin {\theta }_1=u_2\sin {\theta }_2$ ....(i)
as we know,
Time of flight $T=\frac{2u\sin \theta }{g}$
hence,
$\frac{T_1}{T_2}=\frac{\frac{2u_1\sin {\theta }_1}{g}}{\frac{2u_2\sin {\theta }_2}{g}}$
$\frac{T_1}{T_2}=\frac{u_1\sin {\theta }_1}{u_2\sin {\theta }_2}$
From eqn (i)
$\frac{T_1}{T_2}=1$
$T_1=T_2$
Hence, for both the projectiles time of flight will be same.