QJEEM24Ph48c9: JEE Main 2024 Physics - Class 11 JEEM24Ph48c9 Solution: The angle of projection for a projectile to have same…

To find the angle of projection for a projectile to have the same horizontal range and maximum height, we need to express both the range and the maximum height in terms of the projectile's initial velocity and the angle of projection, and then set them equal to each other.

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

$R = \frac{v^2}{g} \sin 2\theta,$

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

The formula for the maximum height $H$ reached by the projectile is:

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

To have the same numerical value for $R$ and $H$ , we set them equal to each other:

$\frac{v^2}{g} \sin 2\theta = \frac{v^2}{2g} \sin^2\theta.$

Simplifying this equation, we get:

$2 \sin 2\theta = \sin^2\theta.$

Using the double-angle formula, $\sin 2\theta = 2 \sin\theta \cos\theta$ , we can rewrite the equation as:

$4 \sin\theta \cos\theta = \sin^2\theta.$

This can be simplified further to:

$4 \sin\theta \cos\theta = (\sin\theta)^2.$

Dividing both sides by $\sin\theta$ (assuming $\sin\theta \neq 0$ ), we get:

$4 \cos\theta = \sin\theta.$

Now, dividing both sides by $\cos\theta$ , we have:

$4 = \tan\theta.$

So, the angle of projection $\theta$ is:

$\theta = \tan^{-1}(4).$

Therefore, the correct answer is:

Option D $\tan ^{-1}(4)$ .