Question

Two particles A & B are projected with same speed so that the ratio of their maximum height reached is 3:1. If the speed of A is doubled without altering other parameters, the ratio of their horizontal ranges is

• A. 1:1
• B. 2:1
• C. 4:1
• D. 3:2

Answer 1

The correct option is C

4:1

From the question, $u_1$ = $u_2$ = u(say)
We know already, H = $\frac{u^{2}si{n}^2\theta }{2g}$
Let angles at which both are projected be ${\theta }_A$ $\&$ ${\theta }_B$
$H_A$ = $\frac{u^{2}si{n}^2{\theta }_A}{2g}$, $H_B$ = $\frac{u^{2}si{n}^2{\theta }_B}{2g}$
Given, $\frac{H_A}{H_B}$ = $\frac{3}{1}${Let $H_A$ > $H_B$}
$\frac{u^{2}si{n}^2{\theta }_A}{u^{2}si{n}^2{\theta }_B}$ = $\frac{3}{1}$
$\frac{sin{\theta }_A}{sin{\theta }_B}$ = $\frac{\sqrt{3}}{1}$
[can I assume sin ${\theta }_A$ = $\frac{1}{\sqrt{3}}$sin ${\theta }_B$$\frac{1}{3}$ will I get the same answer? Check]
If ${\theta }_1$ = ${60}^{\circ }$ $\&$ ${\theta }_2$ = ${30}^{\circ }$
Then $\frac{sin{60}^{\circ }}{sin{30}^{\circ }}$ = $\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$ = $\sqrt{3}$
So the angles are ${60}^{\circ }$ = ${\theta }_1$,${30}^{\circ }$ = ${\theta }_2$
Now $u_1$ = 2u
And we know, R = $\frac{u^{2}sin2\theta }{g}$
$\frac{R_A}{R_B}$ = $\frac{\left(\right(2u{)}^{2}sin2{\theta }_A)1g}{\left(u^{2}sin2{\theta }_B\right)1g}$ = $\frac{sin2{\theta }_A}{4sin2{\theta }_A}$
= $\frac{4sin{120}^{\circ }}{sin{60}^{\circ }}$
= $\frac{4sin({180}^{\circ }-{120}^{\circ })}{sin{60}^{\circ }}$
$\therefore$ sin$\theta$ = sin($\pi$ - $\theta$)
= 4.