Two projectile thrown at 30^{circ} and 45^{ci | vedclass

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Question

Time taken to reach maximum height

$t =\frac{ u \sin 	heta}{ g }$

$	herefore \frac{ u _{1} \sin 	heta_{1}}{ g }=\frac{ u _{2} \sin 	heta_{2}

Vedclass · Accepted Answer

$\sqrt{2}: 1$

Vedclass · Answer

Time taken to reach maximum height

$t =\frac{ u \sin 	heta}{ g }$

$	herefore \frac{ u _{1} \sin 	heta_{1}}{ g }=\frac{ u _{2} \sin 	heta_{2}}{ g }$

$\Rightarrow u _{1} \sin 30= u _{2} \sin 45$

$\Rightarrow \frac{ u _{1}}{ u _{2}}=\frac{1 / \sqrt{2}}{1 / 2}=\frac{\sqrt{2}}{1}$

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

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