Two projectiles are fired from ground with same initial speeds from same point at angles $\left(45^{\circ}+\right.$ $\alpha)$ and $\left(45^{\circ}-\alpha\right)$ with horizontal direction. The ratio of their times of flights is

<p>To find the ratio of the times of flight for two projectiles fired from the ground with the same initial speed but at different angles, we follow these steps:</p> <p><strong>Projectile Angles:</strong></p> <p><p>The first projectile is launched at an angle $\theta_1 = 45^\circ + \alpha$.</p></p> <p><p>The second projectile is launched at an angle $\theta_2 = 45^\circ - \alpha$.</p></p> <p><strong>Time of Flight Formula:</strong></p> <p><p>The time of flight $T$ for a projectile is given by:</p> <p>$ T = \frac{2v \sin \theta}{g} $</p> <p>where $v$ is the initial speed and $g$ is the acceleration due to gravity.</p></p> <p><strong>Ratio of Times of Flight:</strong></p> <p><p>The ratio of the times of flights $\frac{T_1}{T_2}$ can be determined as follows:</p> <p>$ \frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)} $</p></p> <p><strong>Applying Trigonometric Identities:</strong></p> <p><p>Use the sine addition and subtraction formulas:</p> <p>$ \sin(45^\circ + \alpha) = \frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha) $</p> <p>$ \sin(45^\circ - \alpha) = \frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha) $</p></p> <p><strong>Simplifying the Ratio:</strong></p> <p><p>Substitute the expressions for $\sin(45^\circ + \alpha)$ and $\sin(45^\circ - \alpha)$ into the ratio:</p> <p>$ \frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}} (\cos \alpha + \sin \alpha)}{\frac{1}{\sqrt{2}} (\cos \alpha - \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} $</p></p> <p><p>Factor out the trigonometric functions to express in terms of tangent:</p> <p>$ \frac{T_1}{T_2} = \frac{1 + \tan \alpha}{1 - \tan \alpha} $</p></p> <p>Therefore, the ratio of the times of flights for the two projectiles is $\frac{1 + \tan \alpha}{1 - \tan \alpha}$.</p>