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