The equations of motion of a projectile are given by $x=36 t \mathrm{~m}$ and $2 y=96 t-9.8 t^2 \mat

Question

The equations of motion of a projectile are given by $x=36 t \mathrm{~m}$ and $2 y=96 t-9.8 t^2 \mathrm{~m}$. The angle of projection is

Professor PC Thomas Classes · Answer

(a) : Given : $x=36 t, 2 y=96 t-9.8 t^2$ or $y=48 t-4.9 t^2$ Let the initial velocity of projectile be $u$ and angle of projection is $	heta$. Then, Initial horizontal component of velocity, $$ u_x=u \cos 	heta=\left(\frac{d x}{d t}ight)_{t=0}=36 	ext { or } u \cos 	heta=36 $$ Initial vertical component of velocity, $$ u_y=u \sin 	heta=\left(\frac{d y}{d t}ight)_{t=0}=48 	ext { or } u \sin 	heta=48 $$ Dividing (ii) by (i), we get $	an 	heta=\frac{48}{36}=\frac{4}{3} ; \quad 	herefore \quad \sin 	heta=\frac{4}{5} \quad$ or $\quad 	heta=\sin ^{-1}\left(\frac{4}{5}ight)$