The equation for the trajectory of a projectile is $y=\left(\frac{x}{\sqrt{3}}-\frac{x^2}{60}\right) m$. The…
The equation for the trajectory of a projectile is $y=\left(\frac{x}{\sqrt{3}}-\frac{x^2}{60}\right) m$. The velocity of projection of the projectile is (Acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$8 \mathrm{~ms}^{-1}$
$40 \mathrm{~ms}^{-1}$
$16 \mathrm{~ms}^{-1}$
$20 \mathrm{~ms}^{-1}$
Solution
Coefficient of $x^2$ in equation of trajectory
$
=\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}
$
So, $\frac{\mathrm{g}}{2 \mathrm{u}^2 \cos ^2 \theta}=\frac{1}{60} \Rightarrow \frac{10}{2 \mathrm{u}^2 \cos ^2 \theta}=\frac{1}{60}$
$
\Rightarrow \mathrm{u}^2 \cos ^2 \theta=300
$
Coefficient of $x$ in equation of trajectory is $\tan \theta$
So, $\tan \theta=\frac{1}{\sqrt{3}}$
$
\Rightarrow \theta=30^{\circ}
$
As $u^2 \cos ^2 \theta=300$
$
\Rightarrow \mathrm{u}^2\left(\frac{3}{4}\right)=300
$
$
\Rightarrow \mathrm{u}^2=400
$
$\Rightarrow \mathrm{u}=20 \mathrm{~m} / \mathrm{s}$