QJEEM22Ph1a0a: JEE Main 2022 Physics - Class 11 JEEM22Ph1a0a Solution: A ball is projected from the ground with a speed 15 ms 1 at an…

To solve this problem, we will use the equations for the range and maximum height of a projectile. The range $R$ and maximum height $H$ of a projectile launched with speed $u$ at an angle $\theta$ can be expressed as follows:

Range:

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

Maximum height:

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

We are given that the range and maximum height are equal. So, we set these equations equal to each other:

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

We can cancel out the common terms $u^2$ and $g$ on both sides:

$\sin(2\theta) = \frac{1}{2} \sin^2(\theta)$

Using the double angle identity for sine, $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ , we substitute it into the equation:

$2 \sin(\theta) \cos(\theta) = \frac{1}{2} \sin^2(\theta)$

We can simplify this by dividing both sides by $\sin(\theta)$ (assuming $\theta \neq 0$ ):

$2 \cos(\theta) = \frac{1}{2} \sin(\theta)$

Rearranging to get all terms on one side gives us:

$4 \cos(\theta) = \sin(\theta)$

Dividing both sides by $\cos(\theta)$ , we get:

$4 = \tan(\theta)$

So, we find that:

$\tan(\theta) = 4$

Thus, the correct answer is:

Option D: 4