Two projectiles are fired from the ground wit | vedclass

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(D) Let the initial speed be $v$. The angles of projection are $	heta_1 = 45^{\circ} + \alpha$ and $	heta_2 = 45^{\circ} - \alpha$.The formula for t

Vedclass · Accepted Answer

$\frac{1+	an \alpha}{1-	an \alpha}$

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(D) Let the initial speed be $v$. The angles of projection are $	heta_1 = 45^{\circ} + \alpha$ and $	heta_2 = 45^{\circ} - \alpha$.The formula for the time of flight is $T = \frac{2v \sin 	heta}{g}$.Taking the ratio of the times of flight $T_1$ and $T_2$:$\frac{T_1}{T_2} = \frac{\sin(45^{\circ} + \alpha)}{\sin(45^{\circ} - \alpha)}$.Using the trigonometric expansion $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$\frac{T_1}{T_2} = \frac{\sin 45^{\circ} \cos \alpha + \cos 45^{\circ} \sin \alpha}{\sin 45^{\circ} \cos \alpha - \cos 45^{\circ} \sin \alpha}$.Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha)}{\frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha}$.Dividing the numerator and denominator by $\cos \alpha$:$\frac{T_1}{T_2} = \frac{1 + 	an \alpha}{1 - 	an \alpha}$.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

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

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