Question

A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form $y=px-q{x}^2$ where p and q are positive constants. Find the velocity of the particle at the origin of co-ordinates.

Answer 1

Comparing the given equation with the equation of a projectile motion,
$y=x\tan \theta -\frac{g{x}^2}{2u^2}(1+{\tan }^2\theta )$
We find that $g=a,\tan \theta =pand\frac{a}{2u^2}(1+{\tan }^2\theta )=q$
u= velocity of particle at origin
$=\sqrt{\frac{a(1+{\tan }^2\theta )}{2q}}=\sqrt{\frac{a(1+p^2)}{2q}}$