Motion in a Plane

Here, two projectiles are projected at angles 42° and 48° withsame initial velocity. As we know the expression of range of projectile, Range $=\frac{u^{2} \sin 2\theta}{g}$ At $\theta_1$ = 42° Range, $R_{1}=\frac{u^{2} \sin 2(42)^{\circ}}{g}=\frac{0.99 u^{2}}{g}$ At $\theta_2$ = 48° Range, $R_{2}=\frac{u^{2} \sin 2(48)^{\circ}}{g}=\frac{0.99 u^{2}}{g}$ The range of the projectile is same for the two projectiles. Therefore, $R_1 = R_2$ Now, as we know the expression of height of the projectile, $H_{\max}=\frac{u^{2} \sin \theta}{2g}$ At 42°, $H_{\max}=H_{1}=\frac{u^{2} \sin 42^{\circ}}{2g}=\frac{0.669 u^{2}}{2g}$ At 48°, $H_{\max}=H_{2}=\frac{u^{2} \sin 48^{\circ}}{2g}=\frac{0.743 u^{2}}{2g}$ Higher the value of q higher the value of maximum height.Therefore, $H_1 < H_2$.