A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^2}{25 g}$, where value of $n$ is: (Given, ' $g$ ' is the acceleration due to gravity.)

<p>Given that the horizontal range of a projectile is three times its maximum height, we need to determine the value of $ n $ in the expression for the range $\frac{n u^2}{25 g}$.</p> <p>Let's start by setting up the equations for the range and maximum height of a projectile.</p> <p>The formula for the range $ R $ is:</p> <p>$ R = \frac{u^2 \sin 2\theta}{g} $</p> <p>The formula for the maximum height $ H_{\max} $ is:</p> <p>$ H_{\max} = \frac{u^2 \sin^2 \theta}{2g} $</p> <p>According to the problem, the range $ R $ is three times the maximum height $ H_{\max} $:</p> <p>$ R = 3H_{\max} $</p> <p>Substituting the formulas for $ R $ and $ H_{\max} $ into this condition gives:</p> <p>$ \frac{u^2 \sin 2\theta}{g} = 3 \cdot \frac{u^2 \sin^2 \theta}{2g} $</p> <p>Simplifying this equation, we get:</p> <p>$ 2 \sin \theta \cos \theta = \frac{3}{2} \sin^2 \theta $</p> <p>Using the identity $ \sin 2\theta = 2 \sin \theta \cos \theta $, we can rewrite:</p> <p>$ 2 \sin \theta \cos \theta = \sin 2\theta $</p> <p>This gives us:</p> <p>$ \sin 2\theta = \frac{3}{2} \sin^2 \theta $</p> <p>Dividing both sides by $\sin \theta$, assuming $\sin \theta \neq 0$, we have:</p> <p>$ 2 \cos \theta = \frac{3}{2} \sin \theta $</p> <p>Rearranging the terms to solve for $\tan \theta$, we get:</p> <p>$ \tan \theta = \frac{4}{3} $</p> <p>Thus, the angle $\theta$ where $\tan \theta = \frac{4}{3}$ corresponds to $\theta = 53^\circ$.</p> <p>Now, substituting back into the range formula:</p> <p>$ R = \frac{u^2 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}}{g} $</p> <p>The calculation yields:</p> <p>$ R = \frac{24 u^2}{25 g} $</p> <p>Therefore, the value of $ n $ in the expression $\frac{n u^2}{25 g}$ is $ n = 24 $.</p>