If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum height reached by it will

The correct option is Be quadrupled.

To understand why the maximum height reached by the projectile is quadrupled when its initial velocity is doubled, let us analyze the formula for the maximum height of a projectile.

The maximum height ($H$) reached by a projectile launched with an initial velocity $u$ at an angle $\theta$ with the horizontal is given by the formula:

$H=\frac{u^2{\sin }^2\theta }{2g}$

where:
- $u$ is the initial velocity of the projectile,
- $\theta$ is the angle of projection, and
- $g$ is the acceleration due to gravity.

From this formula, we can see that if the angle of projection ($\theta$) and the acceleration due to gravity ($g$) are kept constant, the maximum height is directly proportional to the square of the initial velocity:

$H\propto u^2$

Let the initial maximum height be $H_1$ with initial velocity $u_1=u$:
$H_1=\frac{u^2{\sin }^2\theta }{2g}$

If the initial velocity is doubled, the new velocity becomes $u_2=2u$. The new maximum height $H_2$ is:
$H_2=\frac{{\left(2u\right)}^2{\sin }^2\theta }{2g}$

Simplifying the expression for $H_2$:
$H_2=\frac{4u^2{\sin }^2\theta }{2g}$

Substituting $H_1$ back into the equation:
$H_2=4H_1$

Therefore, if the initial velocity of a projectile is doubled while keeping the angle of projection the same, the maximum height reached by it will be quadrupled (increase by a factor of 4).