Question

The equations of motion of a projectile are given by $x=36t$ metres and $2y=96t-9.8t^2$ metres. The angle of projection is

• A. ${\sin }^{-1}\left(\frac{4}{5}\right)$
• B. ${\sin }^{-1}\left(\frac{3}{5}\right)$
• C. ${\sin }^{-1}\left(\frac{4}{3}\right)$
• D. ${\sin }^{-1}\left(\frac{3}{4}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left(\frac{4}{5}\right)$

$x=36t\therefore v_x=\frac{dx}{dt}=36m/s$
$y=48t-4.9t^2\therefore v_y=48-9.8t$
At $t=0,v_x=36m/s$ and $v_y=48m/s$
So, angle of projection
$\theta ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$
or $\theta ={\sin }^{-1}\left(4/5\right)$