Question 23: Motion in 2D & Vectors

Question

Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :

Accepted Answer

$$\begin{aligned}
& \mathrm{H}_{\mathrm{Max}}=\frac{(\mathrm{usin} 	heta)^2}{2 \mathrm{~g}} \
& \frac{\left(\mathrm{H}_{\max }ight)_1}{\left(\mathrm{H}_{\max }ight)_2}=\frac{\mathrm{u}^2 \sin ^2(45-\alpha)}{\mathrm{u}^2 \sin ^2(45+\alpha)} \
& =\frac{\left(\frac{1}{\sqrt{2}} \cos \alpha-\frac{1}{\sqrt{2}} \sin \alphaight)^2}{\left(\frac{1}{\sqrt{2}} \cos \alpha+\frac{1}{\sqrt{2}} \sin \alphaight)^2} \
& =\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}
\end{aligned}$$

Answer

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

Answer

$ \frac{1+\sin2\alpha}{1-\sin2\alpha} $

Answer

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

Answer

$ \frac{1-\sin2\alpha}{1+\sin2\alpha} $

Motion in 2D & Vectors

Two projectiles are fired with same initial speed from same point on ground at angles of (45^\circ - \alpha) and (45^\circ + \alpha), respectively, with the horizontal direction. The ratio of their maximum heights attained is :

Choose the correct answer: