Question

A particle initially at origin, is moving in x-y
plane such that its velocity component in x direction
remains constant and equal to 3 m/s and its velocity
component in y direction varies with time as vy= 2t
m/s,
where t is time in second, The equation of trajectory of
the particle will be

Answer 1

Well,

$\quad$

$\displaystyle\longrightarrow\sf {u_x =u \cos \theta=3\ m\ s^{-1}}$

$\quad$

$\displaystyle\longrightarrow\sf {u_y=u\sin\theta=2t\ m\ s^{-1}}$

$\quad$

The equation of the trajectory is,

$\quad$

$\displaystyle\longrightarrow\sf {y=x\tan\theta-\dfrac {g}{2(u\cos\theta)^2}x^2}$

$\quad$

$\displaystyle\longrightarrow\sf {y=x\cdot\dfrac {u\sin\theta}{u\cos\theta}-\dfrac {10}{2(3)^2}x^2}$

$\quad$

$\displaystyle\longrightarrow\sf {y=x\cdot\dfrac {2t}{3}-\dfrac {5}{9}x^2\quad\quad\dots (1)}$

$\quad$

But, if horizontal motion is taken into consideration, by second kinematic equation, (assume no external horizontal force is acting)

$\quad$

$\displaystyle\longrightarrow\sf {x=u\cos\theta\cdot t}$

$\quad$

$\displaystyle\longrightarrow\sf {t=\dfrac {x}{u\cos\theta}}$

$\quad$

$\displaystyle\longrightarrow\sf {t=\dfrac {x}{3}}$

$\quad$

Then (1) becomes,

$\quad$

$\displaystyle\longrightarrow\sf {y=x\cdot\dfrac {2}{3}\cdot\dfrac {x}{3}-\dfrac {5}{9}x^2}$

$\quad$

$\displaystyle\longrightarrow\sf {y=\dfrac {2}{9}x^2-\dfrac {5}{9}x^2}$

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {y=-\dfrac {x^2}{3}}}}$

$\quad$

Or,

$\quad$

$\displaystyle\longrightarrow\sf {\underline {\underline {x^2+3y=0}}}$