Question

The equation of motion of a projectile are given by x = 36 t metre and $2y=96t–9.8t^2$ metre. The angle of projection is

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

Answer 1

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

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