A projectile is thrown with an initial velocity \[u = a\hat i + b\hat j\]. If the range of the projectile is doubled, then find the maximum height reached by it.
A. \[b = 2a\]
B. \[b = 3a\]
C. \[b = 12a\]
D. \[b = 4a\]

Hint: Resolve the initial velocity into horizontal and vertical components and use the formula for Range. If $\theta $ is the angle of projection, then for a body having same horizontal range and the maximum height is given by the relation, \[R = 4H\cot \theta \]

Complete step by step solution:
As the initial velocity is given in the vector form, we first find the components of velocity along horizontal and vertical directions. If $\theta $ is the angle of projection, then
The velocity along horizontal direction is given by \[u\cos \theta = a\]
The velocity along horizontal direction is given by \[u\sin \theta = b\]

Dividing the above equations we get\[\tan \theta = \dfrac{{u\sin \theta }}{{u\cos \theta }} = \dfrac{b}{a}\]
If $\theta $ is the angle of projection, then for a body having same horizontal range and the maximum height is given by the relation, \[R = 4H\cot \theta \]
Since it is given that R = 2H
\[ \Rightarrow 4H\cot \theta = 2H \Rightarrow \cot \theta = \dfrac{1}{2} \Rightarrow \tan \theta = 2\]
From above relations we get\[\dfrac{b}{a} = 2 \Rightarrow b = 2a\]

Hence, the correct option is (A).

Note: Projectile is the name given to a body thrown with some initial velocity with the horizontal dimensions under the action of gravity alone. Horizontal range is the distance travelled by a projectile during its flight. The horizontal range is the same when the angle of projection of an object is $\theta $ or $90 - \theta $ with the horizontal direction. In other words horizontal range is the same whether $\theta $ the angle of projection with the horizontal or the vertical is. The key components that we need to remember in order to solve projectile motion problems is the angle \[\theta \] at which a projectile is launched and its initial velocity.