For angles of projection of a projectile at angle $\left(45^{\circ}- heta ight)$ and $\left(45^{\c

Question

For angles of projection of a projectile at angle $\left(45^{\circ}-	hetaight)$ and $\left(45^{\circ}+	hetaight)$ with identical initial speeds, the maximum heights achieved by the projectile are in the ratio of

Professor PC Thomas Classes · Answer

(a) : Maximum height, $H=\frac{u^{2} \sin^{2} \alpha}{2 g}$ \ For angle of projection $\left(45^{\circ}-	hetaight)$, $H_{1}=\frac{u^{2} \sin^{2} \left(45^{\circ}-	hetaight)}{2 g}$ \ For angle of projection $\left(45^{\circ}+	hetaight)$, $H_{2}=\frac{u^{2} \sin^{2} \left(45^{\circ}+	hetaight)}{2 g} = \frac{u^{2} \cos^{2} \left(45^{\circ}-	hetaight)}{2 g}$ (since $\sin(90^\circ - x) = \cos x$) \ $	herefore \quad \frac{H_{1}}{H_{2}}=\frac{\sin^{2}\left(45^{\circ}-	hetaight)}{\cos^{2}\left(45^{\circ}-	hetaight)} = 	an^{2}\left(45^{\circ}-	hetaight)$