Question

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

• A. $si{n}^{-1}\left(\frac{4}{5}\right)$
• B. $si{n}^{-1}\left(\frac{3}{5}\right)$
• C. $si{n}^{-1}\left(\frac{4}{3}\right)$
• D. $si{n}^{-1}\left(\frac{3}{4}\right)$

Answer 1

Answer: (A)

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,
$u_x=ucos\theta ={\left(\frac{dx}{dt}\right)}_{t=0}=36$ or
$ucos\theta =36...\left(i\right)$
Initial vertical component of velocity,
$u_y=usin\theta ={\left(\frac{dy}{dt}\right)}_{t=0}=48$ 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
$\theta =si{n}^{-1}\left(\frac{4}{5}\right)$