### Related Formula Maximum height of a projectile: H_textmax = fracu^2 sin^2 theta2g$H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}$ where theta$\theta$ is the angle of projection measured from the horizontal ground. ### Core Logic The angle of projection in this question is measured from the **vertical axis** as phi$\phi$. This relates to the horizontal projection angle theta$\theta$ as: theta = 90^circ - phi$\theta = 90^{\circ} - \phi$ Substitute this into the maximum height formula: h_m = fracu^2 sin^2 (90^circ - phi)2g = fracu^2 cos^2 phi2g$h_m = \frac{u^2 \sin^2 (90^{\circ} - \phi)}{2g} = \frac{u^2 \cos^2 \phi}{2g}$ h_m(phi) propto cos^2 phi$h_m(\phi) \propto \cos^2 \phi$ ### Step 1: Analyzing Boundary Conditions Let's check the boundary values of h_m$h_m$ for phi in [0, 90^circ]$\phi \in [0, 90^{\circ}]$: - At phi = 0^circ$\phi = 0^{\circ}$ (fired straight up vertically): h_m(0) = fracu^22g = textMaximum height$h_m(0) = \frac{u^2}{2g} = \text{Maximum height}$ - At phi = 90^circ$\phi = 90^{\circ}$ (fired horizontally along the ground): h_m(90^circ) = 0$h_m(90^{\circ}) = 0$ Let's evaluate the behavior of h_m(phi)$h_m(\phi)$: - At small angles phi approx 0$\phi \approx 0$, the slope fracd h_mdphi = -fracu^22g sin(2phi) approx 0$\frac{d h_m}{d\phi} = -\frac{u^2}{2g} \sin(2\phi) \approx 0$. This means the curve starts with a horizontal slope (flat top). - As phi$\phi$ increases towards 90^circ$90^{\circ}$, h_m$h_m$ falls smoothly, reaching 0 at 90^circ$90^{\circ}$ with flat slope again. This sinusoidal falloff perfectly matches **Curve (3)**.

### Pattern Recognition Always read the coordinate origin carefully! Measuring angle from the vertical instead of the horizontal turns sin^2theta$\sin^2\theta$ into cos^2phi$\cos^2\phi$. Checking extremes: vertical shot (0^circ$0^{\circ}$) yields maximum height, while flat shot (90^circ$90^{\circ}$) yields zero height. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane: Projectile Motion