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.