Question

Neglecting the air resistance, the time of flight of a projectile is determined by

• A. $U_{\text{vertical}} $
• B. $U_{\text{horizontal}} $
• C. $ U= U_{\text{vertical}}+U_{\text{horizontal}}^{2} $
• D. $U=\left(U_{\text {vertical }}^{2}+U_{\text {horizontal }}^{2}\right)^{1 / 2}$

Answer 1

Answer: (A)

Let a body be projected at an initial velocity $u$ in a direction making an angle $\theta$ with the horizontal and let it take time $t$ to reach the highest point $P$ of its path.
The vertical velocity of the body at $P$ is zero.
From equation of motion. $v=u-g t$
Putting, $v=v_{y}=0$
and $u=u_{y}=u \sin \theta 0=u \sin \theta-g t$
$\Rightarrow t=\frac{u \sin \theta}{g}$
Hence, time of flight $(T)$ is $2 t$.
$\therefore T=2 t=\frac{2 u \sin \theta}{g}=\frac{2}{g} \times U_{\text {vertical }}$