The position of a projectile launched from the origin at t = 0 is given by r = ( 40 i + 50 j),m at t = 2,s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10, ms-2)

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JEE Mains · Physics · STD 11 - Motion in plane

JEE Mains 2014
Hard
MCQ

The position of a projectile launched from the origin at \(t = 0\) is given by \(\vec r = \left( {40\hat i + 50\hat j} \right)\,m\) at \(t = 2\,s\). If the projectile was launched at an angle \(\theta\) from the horizontal, then \(\theta\) is (take \(g = 10\, ms^{-2}\))

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(C) \({\tan ^{ - 1}}\frac{7}{4}\)

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\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,From\,question,\\
\,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,{u_x} = \frac{{40}}{2} = 20m/s\\
\,\,\,\,\,\,\,\,\,Vertical\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,\,50 = {u_y}t + \frac{1}{2}g{t^2}\\
\Rightarrow \,{u_y} \times 2 + \frac{1}{2}\left( { - 10} \right) \times 4\\
or,\,50 = 2{u_y} - 20
\end{array}\)

\(\begin{array}{l}
or,\,\,{u_y} = \frac{{70}}{2} = 35m/s\\
\therefore \,\tan \,\theta = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4}\\
\Rightarrow \,\,Angle\,\theta {\tan ^{ - 1}}\frac{7}{4}
\end{array}\)

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