Question

A particle is projected from the origin in the $x-y$ plane. The acceleration of particle in negative $y-$direction is a. If equation of path of the particle is y $y=ax-b{x}^2$, 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. $\sqrt{\frac{\alpha b}{a^2}}$

Answer 1

Answer: (B)

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

$\left(B\right)$ (There is correction needed for answer. ${}^{\prime }{\infty }^{\prime }$ instead of ${}^{\prime }{a}^{\prime }$)

$y=ax-{bx}^2$

$y=0\Rightarrow ax-{bx}^2=0$

$\therefore$ $x=0$ & $x=\frac{a}{b}$

$\therefore$ Range $R=\frac{a}{b}\to \left(1\right)$

We have $R=\frac{u^2\sin 2\theta }{a}=\frac{2u^2\sin \vartheta cos\vartheta }{a}\to \left(2\right)$

$\left(1\right)=\left(2\right)$ $\Rightarrow \frac{2u^2\sin \theta \cos \theta }{a}=\frac{a}{b}\to \left(3\right)$

$\frac{dy}{dx}=0\Rightarrow x=\frac{a}{2b}$

and $\frac{d^{2}y}{{dx}^2}<0$ $\therefore$ $y_{max}$ is at $x=\frac{a}{2b}$

$\therefore$ $y_{max}=a.\frac{a}{2b}-b.\frac{a^2}{{4b}^2}=\frac{a^2}{4b}\to \left(4\right)$

We have maxi height, $y_{max}=\frac{u^2{\sin }^2\theta }{2a}\to \left(5\right)$

$\left(4\right)=\left(5\right)$ $\Rightarrow \frac{u^2{\sin }^2\theta }{2a}=\frac{a^2}{4b}\to \left(6\right)$

$\left(3\right)/\left(6\right)$ $\Rightarrow \frac{2\cos \theta .2}{sin\theta }=\frac{4}{a}$

$\Rightarrow \tan \theta =\frac{a}{1}$

$\therefore$ $sin\theta =\frac{a}{\sqrt{a^2+1}}\to \left(7\right)$

$\left(7\right)$ in $\left(6\right)$

$\Rightarrow \frac{u^2.\frac{a^2}{a^2+1}}{2a}=\frac{a^2}{4b}$

$\Rightarrow u^2=\frac{2a.(a^2+1)}{4b}$

$\Rightarrow u^2=\frac{a.(a^2+1)}{2b}$

$\Rightarrow u=\sqrt{\frac{a.(a^2+1)}{2b}}$