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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 $g = 10\, ms^{-2}$)
${\tan ^{ - 1}}\frac{2}{3}$
${\tan ^{ - 1}}\frac{3}{2}$
${\tan ^{ - 1}}\frac{7}{4}$
${\tan ^{ - 1}}\frac{4}{5}$
Solution
$\begin{array}{l}
\,\,\,\,\,\,\,\,\,From\,question,\\
\,\,\,\,\,\,\,\,\,Horizontal\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,{u_x} = \frac{{40}}{2} = 20m/s\\
\,\,\,\,\,\,\,\,\,Vertical\,velocity\,\left( {initial} \right),\\
\,\,\,\,\,\,\,\,\,\,50 = {u_y}t + \frac{1}{2}g{t^2}\\
\Rightarrow \,{u_y} \times 2 + \frac{1}{2}\left( { – 10} \right) \times 4\\
or,\,50 = 2{u_y} – 20
\end{array}$
$\begin{array}{l}
or,\,\,{u_y} = \frac{{70}}{2} = 35m/s\\
\therefore \,\tan \,\theta = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4}\\
\Rightarrow \,\,Angle\,\theta {\tan ^{ – 1}}\frac{7}{4}
\end{array}$
Similar Questions
Match the columns
| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
