JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2013
- Medium
- MCQ
A projectile is given an initial velocity of \((\hat i + 2\hat j)\ ms^{-1}\), where \(\hat i\) is along the ground and \(\hat j\) is along the vertical. If \(g = 10\ m/s^2\) , the equation of its trajectory is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(y= 2x- 5x^2\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\vec u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \Rightarrow u\cos \theta = 1,\\
u\sin \theta = 2\\
y = x\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}}\\
\therefore \,y = 2x - \frac{1}{2}g{x^2} = 2x - 5{x^2}
\end{array}\)
\vec u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \Rightarrow u\cos \theta = 1,\\
u\sin \theta = 2\\
y = x\tan \theta - \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}}\\
\therefore \,y = 2x - \frac{1}{2}g{x^2} = 2x - 5{x^2}
\end{array}\)
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