JEE MAIN - Physics (2013 (Offline) - No. 27)
A projectile is given an initial velocity of $$\left( {\widehat i + 2\widehat j} \right)$$ m/s, where $${\widehat i}$$ is along the ground and $${\widehat j}$$ is along the
vertical. If g = 10 m/s2, the equation of its trajectory is:
y = x - 5x2
y = 2x - 5x2
4y = 2x - 5x2
4y = 2x - 25x2
Explanation
$$\overrightarrow u = \widehat i + 2\widehat j = {u_x}\widehat i + {u_y}\widehat j$$
$$ \Rightarrow u\cos \theta = 1,u\sin \theta = 2$$
Also $$x = {u_x}t$$ and
$$y = {u_y}t - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {u_x^2}}$$
$$\therefore$$ $$y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}$$
$$ \Rightarrow u\cos \theta = 1,u\sin \theta = 2$$
Also $$x = {u_x}t$$ and
$$y = {u_y}t - {1 \over 2}g{t^2}$$
$$ \Rightarrow $$ $$y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {u_x^2}}$$
$$\therefore$$ $$y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}$$
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