The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is

The correct answer is √3u² / 2g.

Step-by-step Derivation:

Let the projectile be projected with an initial velocity $u$ at an angle of projection $\theta$ with the horizontal plane.

At the maximum height of its trajectory, the vertical component of the velocity of the projectile becomes zero. Therefore, the velocity of the projectile at the maximum height is purely horizontal and is given by:

$v=u\cos \theta$

According to the given problem, the velocity at the maximum height is half of its initial velocity $u$:

$u\cos \theta =\frac{u}{2}$

Simplifying the equation by dividing both sides by $u$, we get:

$\cos \theta =\frac{1}{2}$

Since the cosine of 60° is equal to 1/2, the angle of projection is:

$\theta =\mathrm{60°}$

The horizontal range $R$ of a projectile on a horizontal plane is given by the formula:

$R=\frac{u^2\sin \left(2\theta \right)}{g}$

where $g$ is the acceleration due to gravity.

Now, substitute the value of $\theta =\mathrm{60°}$ into the horizontal range formula:

$R=\frac{u^2\sin (2\times \mathrm{60°})}{g}$

$R=\frac{u^2\sin \left(\mathrm{120°}\right)}{g}$

Using the trigonometric identity, we can find the value of $\sin \left(\mathrm{120°}\right)$:

$\sin \left(\mathrm{120°}\right)=\sin (\mathrm{180°}-\mathrm{60°})=\sin \left(\mathrm{60°}\right)=\frac{\sqrt{3}}{2}$

Substitute this value back into the range equation:

$R=\frac{u^2\times \frac{\sqrt{3}}{2}}{g}$

$R=\frac{\sqrt{3}{u}^2}{2g}$

Thus, the horizontal range of the projectile is $\frac{\sqrt{3}{u}^2}{2g}$.