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3-2.Motion in Plane
medium
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
A$y= x- 5x^2$
B$y= 2x- 5x^2$
C$4y= 2x- 5x^2$
D$4y= 2x- 25x^2$
(JEE MAIN-2013)
Solution
$\begin{array}{l}
\vec u = \hat i + 2\hat j = {u_x}\hat i + {u_y}\hat j \Rightarrow u\cos \theta = 1,\\
u\sin \theta = 2\\
y = x\tan \theta – \frac{1}{2}\frac{{g{x^2}}}{{u_x^2}}\\
\therefore \,y = 2x – \frac{1}{2}g{x^2} = 2x – 5{x^2}
\end{array}$
\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}$
Standard 11
Physics
