JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Kinematics

With the projection angle $\varphi$ measured from the vertical, the angle above the horizontal is $90^{\circ }-\varphi$, so the vertical component of the launch speed is $u\cos \varphi$. The maximum height reached is

Thus $h_m$ is greatest at $\varphi =0$, a purely vertical throw, and falls to zero at $\varphi =90^{\circ }$, a horizontal throw. Since $\frac{d{h}_m}{d\varphi }=-\frac{u^2}{2g}\sin 2\varphi$ vanishes at $\varphi =0$, the curve leaves its maximum with a horizontal tangent and then decreases to zero at $90^{\circ }$. The graph that represents $h_m(\varphi )$ is the one starting at its maximum value with a flat tangent at $\varphi =0$ and decreasing to zero at $\varphi =90^{\circ }$.