Two projectiles are fired with same initial speed from same point on ground at angles of (45° – ) and (45° + ) respectively, with the horizontal direction. The ratio of their maximum heights attained is [2025]

Question

Two projectiles are fired with same initial speed from same point on ground at angles of (45° – $α$ ) and (45° + $α$ ) respectively, with the horizontal direction. The ratio of their maximum heights attained is [2025]

Smart Achievers · Accepted Answer

(3)

$H_{Max} = \frac{{(u \sin θ)}^{2}}{2 g}$

$= \frac{{(H_{\max})}_{1}}{{(H_{\max})}_{2}} = \frac{u^{2} \sin^{2} (45 - α)}{u^{2} \sin^{2} (45 + α)}$

$= \frac{{(\frac{1}{\sqrt{2}} \cos α - \frac{1}{\sqrt{2}} \sin α)}^{2}}{{(\frac{1}{\sqrt{2}} \cos α + \frac{1}{\sqrt{2}} \sin α)}^{2}} = \frac{1 - \sin 2 α}{1 + \sin 2 α}$ .

Solution

Q.
Two projectiles are fired with same initial speed from same point on ground at angles of (45° – $α$ ) and (45° + $α$ ) respectively, with the horizontal direction. The ratio of their maximum heights attained is [2025]

1 $\frac{1 - \tan α}{1 + \tan α}$

2 $\frac{1 + \sin α}{1 - \sin α}$

3 $\frac{1 - \sin 2 α}{1 + \sin 2 α}$

4 $\frac{1 + \sin 2 α}{1 - \sin 2 α}$

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Solution

Q. Two projectiles are fired with same initial speed from same point on ground at angles of (45° – α) and (45° + α) respectively, with the horizontal direction. The ratio of their maximum heights attained is [2025]

1 1–tan α1+tan α

2 1+sin α1–sin α

3 1–sin 2α1+sin 2α

4 1+sin 2α1–sin 2α