Question

A particle is projected from the origin in $xy$ plane. Acceleration of particle in $y$ direction is $\alpha$ . Its equation of path of projectile is $y = ax-bx^2$, then initial velocity of the particle is

• A. $\sqrt{\frac{\alpha}{2b}}$
• B. $\sqrt{\frac{\alpha(1+a^2)}{2b}}$
• C. $\sqrt{\frac{\alpha}{a^2}}$
• D. $\sqrt{\frac{\alpha\,a^2}{b}}$

Answer 1

Answer: (B)

$\sqrt{\frac{\alpha(1+a^2)}{2b}}$

Answer 2

y = ax-bx$^2$ $y = x$ tan $\theta - \frac{\alpha \,x^2}{2 u^2 \, \cos^2 \, \theta}$ $\because$ tan $\theta$ = a, $\frac{\alpha}{2u^2 \, \cos^2 \, \theta} = b;$ $\because \, u = \sqrt{\frac{\alpha (1 + a^2)}{2b}}$