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ઉદગમ સ્થાનથી $t = 0$ સમયે ફેંકેલ પદાર્થ નું સ્થાન $t = 2\,s$ સમયે $\vec r = \left( {40\hat i + 50\hat j} \right)\,m$ છે. જો પદાર્થને સમક્ષિતિજથી $\theta$ કોણે ફેંકવામાં આવ્યો હોય તો $\theta$ શું હશે? ($g = 10\, ms^{-2}$)
${\tan ^{ - 1}}\frac{2}{3}$
${\tan ^{ - 1}}\frac{3}{2}$
${\tan ^{ - 1}}\frac{7}{4}$
${\tan ^{ - 1}}\frac{4}{5}$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,From\,question,\\
\,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,{u_x} = \frac{{40}}{2} = 20m/s\\
\,\,\,\,\,\,\,\,\,Vertical\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,\,50 = {u_y}t + \frac{1}{2}g{t^2}\\
\Rightarrow \,{u_y} \times 2 + \frac{1}{2}\left( { – 10} \right) \times 4\\
or,\,50 = 2{u_y} – 20
\end{array}$
$\begin{array}{l}
or,\,\,{u_y} = \frac{{70}}{2} = 35m/s\\
\therefore \,\tan \,\theta = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4}\\
\Rightarrow \,\,Angle\,\theta {\tan ^{ – 1}}\frac{7}{4}
\end{array}$
