A particle is projected with velocity u$u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as fracnu^225g$\frac{nu^2}{25g}$ , where value of n$n$ is: (Given ' g$g$' is the acceleration due to gravity).

### Related Formula For a projectile with launch speed u$u$ and angle theta$\theta$: - Horizontal Range: R = fracu^2 sin(2theta)g = frac2 u^2 sintheta costhetag$$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}$$ - Maximum Height: H = fracu^2 sin^2theta2g$$H = \frac{u^2 \sin^2\theta}{2g}$$ The general ratio linking range and maximum height is: tantheta = frac4HR$$\tan\theta = \frac{4H}{R}$$ ### Core Logic Given state: R = 3H Rightarrow fracHR = frac13$$R = 3H \Rightarrow \frac{H}{R} = \frac{1}{3}$$ ### Step 1: Determine the projection angle (theta$\theta$) Substitute the ratio into the relation: tantheta = 4 left(fracHRright) = 4 left(frac13right) = frac43$$\tan\theta = 4 \left(\frac{H}{R}\right) = 4 \left(\frac{1}{3}\right) = \frac{4}{3}$$ This is a standard Pythagorean triangle angle: sintheta = frac45, quad costheta = frac35$$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}$$ ### Step 2: Compute the horizontal range (R$R$) R = frac2 u^2 sintheta costhetag$$R = \frac{2 u^2 \sin\theta \cos\theta}{g}$$ R = frac2 u^2 left(frac45right) left(frac35right)g = frac24 u^225 g$$R = \frac{2 u^2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right)}{g} = \frac{24 u^2}{25 g}$$ Comparing this with the given format fracnu^225g$\frac{nu^2}{25g}$: n = 24$n = 24$ ### Pattern Recognition The relation tantheta = 4H/R$\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever R = k H$R = k H$, then tantheta = 4/k$\tan\theta = 4/k$. Recognizing standard angles like tantheta = 4/3$\tan\theta = 4/3$ or 3/4$3/4$ directly yields trigonometric values immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane