QJEEM22Ph4971: JEE Main 2022 Physics - Class 11 JEEM22Ph4971 Solution: Given below are two statements. One is labelled as Assertion A…

When two projectiles are thrown with the same initial velocity 'u' but at complementary angles (say, $\theta$ and $(90^\circ - \theta)$ ) with the horizontal, they attain the same range R. The formula for the range R of a projectile is:

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

For complementary angles, $2 \theta$ and $180^\circ - 2 \theta$ (which simplifies to the same value for the sine function), the ranges are equal.

The maximum height $h_1$ for angle $\theta$ is given by:

$h_1 = \frac{u^2 \sin^2 \theta}{2g}$

And the maximum height $h_2$ for angle $(90^\circ - \theta)$ is given by:

$h_2 = \frac{u^2 \cos^2 \theta}{2g}$

Now, multiplying these heights:

$h_1 h_2 = \left( \frac{u^2 \sin^2 \theta}{2g} \right) \cdot \left( \frac{u^2 \cos^2 \theta}{2g} \right)$

This simplifies to:

$h_1 h_2 = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2}$

Since $\sin^2 \theta \cos^2 \theta = \left( \frac{\sin 2 \theta}{2} \right)^2 = \frac{1}{4} \sin^2 2 \theta$ :

$h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{16g^2}$

Using the range formula $R = \frac{u^2 \sin 2 \theta}{g}$ , we get:

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

Thus, we have:

$4h_1 h_2 = \frac{u^4 \sin^2 2 \theta}{4g^2} = \frac{R^2}{4}$

This simplifies to:

$R = 4\sqrt{h_1 h_2}$

Both the assertion and the reason are correct, and the reason correctly explains the assertion.

The correct answer is:

Option A : Both A and R are true and R is the correct explanation of A.