Two projectile thrown at 30^{circ} and 45^{circ} with horizontal respectively, reach maximum height

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3-2.Motion in Plane

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Two projectile thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

A

$1: \sqrt{2}$

B

$2: 1$

C

$\sqrt{2}: 1$

D

$1: 2$

(JEE MAIN-2022)

Solution

Time taken to reach maximum height

$t =\frac{ u \sin \theta}{ g }$

$\therefore \frac{ u _{1} \sin \theta_{1}}{ g }=\frac{ u _{2} \sin \theta_{2}}{ g }$

$\Rightarrow u _{1} \sin 30= u _{2} \sin 45$

$\Rightarrow \frac{ u _{1}}{ u _{2}}=\frac{1 / \sqrt{2}}{1 / 2}=\frac{\sqrt{2}}{1}$

Standard 11

Physics

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