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

$\overrightarrow{v}=a \hat{i}+b \hat{j} \Rightarrow u \cos \theta =a \text{and} u \sin \theta =b$

$$\begin{gathered} \tan \theta =\frac{b}{a} \\ R=2H_{max} \\ \text{also} \frac{u^2\sin 2\theta }{g}=\frac{2u^2{\sin }^2\theta }{2g} \\ \Rightarrow \tan \theta =2 \\ \therefore \frac{b}{a}=2 \Rightarrow b=2a \end{gathered}$$