The equation of motion of a projectile are gi | vedclass

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Question

(a) $x = 36t$

${v_x} = \frac{{dx}}{{dt}} = 36\,m/s$

$y = 48t - 4.9{t^2}$

${v_y} = 48 - 9.8t$

at $t = 0$ ${v_x} = 36$ and ${v_y} = 48\,m/s$

Vedclass · Accepted Answer

${\sin ^{ - 1}}\left( {\frac{4}{5}} \right)$

Vedclass · Answer

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

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

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