Question

A projectile is thrown with an initial velocity of $\overrightarrow{v}=a\hat{i}+b\hat{j}$. If the range of projectile is double the maximum height reached by it. Find the ratio $\frac{b}{a}$

• A. $\frac{b}{a}=2$
• B. $\frac{b}{a}=\frac{1}{2}$
• C. $\frac{b}{a}=4$
• D. $\frac{b}{a}=\frac{1}{4}$

Answer 1

The correct option is C $\frac{b}{a}=4$

$\because \overrightarrow{v}=a\hat{i}+b\hat{j}vsin\theta =bvcos\theta =aR=\frac{u^{2}sin2\theta }{g}=\frac{2\ast usin\theta \ast ucos\theta }{g}=\frac{2ab}{g}H=\frac{u^{2}si{n}^2\theta }{2g}=\frac{b^2}{2g}R=2H\frac{2ab}{g}=2\frac{b^2}{2g}b=2a$