2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The position of a projectile launched from the origin at t = 0is given by \[\vec{r}=\left( 40\hat{i}+50\hat{j} \right)m\]at \[t=2s.\]If the projectile was launched at an angle \[\theta \] from the horizontal, then \[\theta \] is (take \[g=10\,m{{s}^{-2}}\])   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) \[{{\tan }^{-1}}\frac{2}{3}\]                        

    B) \[{{\tan }^{-1}}\frac{3}{2}\]

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

    D) \[{{\tan }^{-1}}\frac{4}{5}\]

    Correct Answer: C

    Solution :

    From question, Horizontal velocity (initial), \[{{u}_{x}}=\frac{40}{2}=20m/s\] Vertical velocity (initial), \[50={{u}_{y}}t+\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]\[{{u}_{y}}\times 2+\frac{1}{2}(-10)\times 4\]or\[50=2{{u}_{y}}-20\] or\[{{u}_{y}}=\frac{70}{2}=35m/s\] \[\therefore \]\[\tan \theta =\frac{{{u}_{y}}}{{{u}_{x}}}=\frac{35}{20}=\frac{7}{4}\]\[\Rightarrow \]Angle\[\theta ={{\tan }^{-1}}\frac{7}{4}\] \[\therefore \]\[\frac{\sin {{\theta }_{c}}}{\sin {{\theta }_{s}}}=\frac{\frac{5}{7}g}{\frac{2}{3}g}=\frac{15}{14}\]                


You need to login to perform this action.
You will be redirected in 3 sec spinner