Question

Ratio of minimum kinetic energies of two projectiles of same mass is $4:1$. The ratio of the maximum height attained by them is also $4:1$. The ratio of their ranges would be

• A. $16:1$
• B. $4:1$
• C. $8:1$
• D. $2:1$

Answer 1

The correct option is B $4:1$

Velocity of a projectile motion is minimum at the highest point of the trajectory
Thus, $v_{min}=u\cos \theta$
Minimum kinetic of the projectile = $K.E=\frac{1}{2}m(u\cos \theta {)}^2$
According to question, $\begin{array}{}\frac{K.E_1}{K.E_2}=\frac{\frac{1}{2}m{u}_1^2{\cos }^2{\theta }_1}{\frac{1}{2}m{u}_2^2{\cos }^2{\theta }_2}=\frac{4}{1}& \text{(1)}\end{array}$
Let $H_1$ and $H_2$ be the maximum heights of two projectiles. We are given that $\frac{H_1}{H_2}=\frac{4}{1}$
$\begin{array}{}\Rightarrow \frac{\frac{u_1^2{\sin }^2{\theta }_1}{2g}}{\frac{u_2^2{\sin }^2{\theta }_2}{2g}}=\frac{4}{1}& \text{(2)}\end{array}$
Also, we know that $R=\frac{u^2\sin 2\theta }{g}=\frac{2u^2\sin \theta \cos \theta }{g}$
$\Rightarrow \frac{R_1^2}{R_2^2}=\frac{u_1^4{\sin }^2{\theta }_1{\cos }^2{\theta }_1}{u_2^4{\sin }^2{\theta }_2{\cos }^2\theta }$
Multiplying (1) and (2), we get
$\frac{u_1^4{\sin }^2{\theta }_1{\cos }^2{\theta }_1}{u_1^4{\sin }^2{\theta }_2{\cos }^2{\theta }_2}=\frac{16}{1}\Rightarrow \frac{R_1^2}{R_2^2}=\frac{16}{1}\therefore \frac{R_1}{R_2}=\frac{4}{1}$