Question

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

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

Answer 1

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

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

Method II:
Comparing with the standard equations for positions x and y of a projectile we get,
$v_x$ = 36 and $v_y$ = 48 m/s
So, angle of projection $\theta$ = ${tan}^{-1}$($\frac{v_y}{v_x}$) = ${tan}^{-1}$($\frac{4}{3}$)
Or $\theta$ = ${sin}^{-1}$($\frac{4}{5}$)