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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 $\theta_0$ with speed $v_0$ then $(g = 10\, ms^{-2}$)
${\theta _0} = {\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} \right)$ and ${v_0} = \frac{5}{3}\,m{s^{ - 1}}$
${\theta _0} = {\cos ^{ - 1}}\,\left( {\frac{2}{{\sqrt 5 }}} \right)$ and ${v_0} = \frac{3}{5}\,m{s^{ - 1}}$
${\theta _0} = {\sin ^{ - 1}}\,\left( {\frac{2}{{\sqrt 5 }}} \right)$ and ${v_0} = \frac{3}{5}\,m{s^{ - 1}}$
${\theta _0} = {\sin ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} \right)$ and ${v_0} = \frac{5}{3}\,m{s^{ - 1}}$
Solution
$\begin{array}{l}
Equation\,of\,trajectory\,is\,given\,as\\
\,\,\,\,\,\,\,\,\,\,y = 2x – 9{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( A \right)\\
\,\,\,Comparing\,with\,equation\,\,\,\,\,\,\,\,\,\,\,\,\\
\,\,\,\,\,\,\,\,\,y = \,X\,\tan \,\theta – \frac{g}{{2{u^2}{{\cos }^2}\theta }} \cdot {x^2}\,\,\,\,…\left( B \right)\\
\,\,\,\,\,We\,get,\,\tan \,\theta = 2\\
\,\,\,\,\,\,\therefore \,\cos \,\theta = \frac{1}{{\sqrt 5 }}
\end{array}$
$\begin{array}{l}
Also,\,\frac{g}{{2{u^2}{{\cos }^2}\theta }} = 9\\
\Rightarrow \,\,\frac{{10}}{{2 \times 9 \times {{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2}}} = {u^2}\,;\,{u^2} = \frac{{25}}{g}\\
\Rightarrow \,u = \frac{5}{3}\,m/s
\end{array}$
