Question

A particle moves according to the equation $y=ax-b{x}^2$. Only gravitation field is present. The initial magnitude of velocity is given as $v_{initial}=\sqrt{\frac{g}{kb}(1+a^2)}$. Find $k$. (take gravitational acceleration $=g$)

Answer 1

As only gravitational field is present so $a_x=0$ and $a_y=-g$
now $y=ax-b{x}^2$
differentiating with respect to time
$\frac{dy}{dt}=a\frac{dx}{dt}-2bx\frac{dx}{dt}$
or $\frac{dy}{dt}{|}_{x=0}=a\frac{dx}{dt}....\left(1\right)$
and $\frac{d^{2}y}{d{t}^2}=a\frac{d^{2}x}{d{t}^2}-2bx\frac{d^{2}x}{d{t}^2}-2b(\frac{dx}{dt}{)}^2.....(2)$
since $a_x=\frac{d^{2}x}{d{t}^2}=0$ so (2) becomes
$\frac{d^{2}y}{d{t}^2}=-2b(\frac{dx}{dt}{)}^2$
or $-g=-2b(\frac{dx}{dt}{)}^2$
or $v_x=\frac{dx}{dt}=\sqrt{\frac{g}{2b}}$
and using (1), $v_y=\frac{dy}{dt}=a\sqrt{\frac{g}{2b}}$
thus, $v_{initial}=\sqrt{v_x^2+v_y^2}=\sqrt{\frac{g}{2b}(1+a^2)}$

Hence value of $k=2$