Question

Two particles $A$ and $B$ are projected with same speeds so that ratio of their maximum heights reached is $3:1$. If the speed of $A$ is doubled without altering other parameters, the ratio of horizontal ranges attained by $A$ and $B$ is:

• A. $1:1$
• B. $2:1$
• C. $4:1$
• D. $3:2$

Answer 1

The correct option is C $4:1$

$\frac{H_1}{H_2}=\frac{4^2\frac{{\sin }^2{\theta }_1}{25}}{4^2\frac{{\sin }^2{\theta }_1}{25}}=\frac{3}{1}$
$\frac{{\sin }^2{\theta }_1}{{\sin }^2{\theta }_2}=\frac{3}{1}$.
$\frac{\sin {\theta }_1}{\sin {\theta }_2}=\sqrt{3}$
$\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$
Now $v_1=2u$
$\frac{R_1}{R_2}=\frac{(2u{)}^2\frac{\sin 2{\theta }_1}{2}}{4^2\frac{\sin {20}_2}{g}}=\frac{4\times \sin {120}^o}{\sin {60}^o}$
$=4x\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}$
$=4$
$\frac{R_1}{R_2}=4:1$