Question

The trajectory of a projectile near the surface of the earth is given as $y=2x−9x^2$. If it were launched at an angle ${θ}_0$ with speed ${ν}_0$ then ($g=10m{s}^{−2}$) :

• A. ${θ}_0={\cos }^{−1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0=\frac{5}{3}m{s}^{−1}$
• B. ${θ}_0={\sin }^{−1}\left(\frac{1}{\sqrt{5}}\right)$ and $v_0=\frac{5}{3}m{s}^{−1}$
• C. ${θ}_0={\sin }^{−1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0=\frac{3}{5}m{s}^{−1}$
• D. ${θ}_0={\cos }^{−1}\left(\frac{2}{\sqrt{5}}\right)$ and $v_0=\frac{3}{5}m{s}^{−1}$

Answer 1

Answer: (A)

Equation of trajectory is given as $y=2x−9x^2$ ......(1)
Comparing with equation :
$y=x\tan θ−\frac{g}{2u^2{\cos }^2θ}.x^2$ .....(2)
We get;
$\tan θ=2$
$∴\cos θ=\frac{1}{\sqrt{5}}$
Also, $\frac{g}{2u^2{\cos }^2θ}=9$
$⇒\frac{10}{2×9×{\left(\frac{1}{\sqrt{5}}\right)}^2}=u^2$
$⇒u^2=\frac{25}{9}$
$⇒u=\frac{5}{3}m/s$