The trajectory of a projectile near the surfa | vedclass

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Question

$\begin{array}{l}
Equation\,of\,trajectory\,is\,given\,as\
\,\,\,\,\,\,\,\,\,\,y = 2x - 9{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\

Vedclass · Accepted Answer

${	heta _0} = {\cos ^{ - 1}}\,\left( {\frac{1}{{\sqrt 5 }}} ight)$ and ${v_0} = \frac{5}{3}\,m{s^{ - 1}}$

Vedclass · Answer

$\begin{array}{l}
Equation\,of\,trajectory\,is\,given\,as\
\,\,\,\,\,\,\,\,\,\,y = 2x - 9{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( A ight)\
\,\,\,Comparing\,with\,equation\,\,\,\,\,\,\,\,\,\,\,\,\
\,\,\,\,\,\,\,\,\,y = \,X\,	an \,	heta  - \frac{g}{{2{u^2}{{\cos }^2}	heta }} \cdot {x^2}\,\,\,\,...\left( B ight)\
\,\,\,\,\,We\,get,\,	an \,	heta  = 2\
\,\,\,\,\,\,	herefore \,\cos \,	heta  = \frac{1}{{\sqrt 5 }}
\end{array}$

$\begin{array}{l}
Also,\,\frac{g}{{2{u^2}{{\cos }^2}	heta }} = 9\
 \Rightarrow \,\,\frac{{10}}{{2 	imes 9 	imes {{\left( {\frac{1}{{\sqrt 5 }}} ight)}^2}}} = {u^2}\,;\,{u^2} = \frac{{25}}{g}\
 \Rightarrow \,u = \frac{5}{3}\,m/s
\end{array}$

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 $(g = 10\, ms^{-2}$)

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