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Q.
The equations of motion of a projectile are given by $x = 36t\, m$ and $2y = 96t - 9.8t^2\,m$. The angle of projection is
Motion in a Plane
Solution:
Given :
$x = 36t$,
$2y = 96t - 9.8t^2$
or $y = 48t-4.9t^2$
Let the initial velocity of projectile be $u$ and angle of projection is $\theta$. Then,
Initial horizontal component of velocity,
$u_{x}=u\,cos\,\theta=\left(\frac{dx}{dt}\right)_{t = 0}=36$ or
$ucos\theta = 36\,...\left(i\right)$
Initial vertical component of velocity,
$u_{y}=u\,sin\,\theta=\left(\frac{dy}{dt}\right)_{t=0}=48$ or
$usin\theta = 48\,... \left(ii\right)$
Dividing $\left(ii\right)$ by $\left(i\right)$, we get
$tan\,\theta=\frac{48}{36}=\frac{4}{3}\,$
$\therefore sin\,\theta=\frac{4}{5}$ or
$\theta=sin^{-1}\left(\frac{4}{5}\right)$