Question

Trajectories of two projectiles are shown in figure. Let $T_{1}$ and $T_{2}$ be the time of flight and $u_{1}$ and $u_{2}$ are their speeds of projections, then

• A. $T_{2} >\, T_{1}$
• B. $T_{2} = T_{1}$
• C. $u_{1} > \, u_{2}$
• D. $u_{1} < \, u_{2}$

Answer 1

Answer: (B)

Since the height of both the projectiles are same, we have: $H _{1}= H _{2}$
$\frac{ u _{1}^{2} \sin ^{2} \theta_{1}}{2 g }=\frac{ u _{2}^{2} \sin ^{2} \theta_{2}}{2 g }$
$u _{1}^{2} \sin ^{2} \theta_{1}= u _{2}^{2} \sin ^{2} \theta_{2}$
$u _{1} \sin \theta_{1}= u _{2} \sin \theta_{2}$ wher $u$ is initial speed and $\theta$ is the angle of projectile.
$\frac{ T _{1}}{ T _{2}}=\frac{\frac{2 u _{1} \sin \theta_{1}}{ g }}{\frac{2 u _{2} \sin \theta_{2}}{ g }}$
$
=\frac{2 u _{1} \sin \theta_{1}}{ g } \times \frac{ g }{2 u _{2} \sin \theta_{2}}
$
$
=\frac{ u _{1} \sin \theta_{1}}{ u _{2} \sin \theta_{2}}=1
$
$
T _{1}= T _{2}
$