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. $v=u-gt$
Putting, $v=v_y=0$
and $u=u_y=u\sin \theta 0=u\sin \theta -gt$
$\Rightarrow t=\frac{u\sin \theta }{g}$
Hence, time of flight $(T)$ is $2t$.
$\therefore T=2t=\frac{2u\sin \theta }{g}=\frac{2}{g}\times U_{\text{vertical }}$