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Standard XII

Physics

Motional EMF in an Open Conductor

Two projectil...

Question

Two projectiles are projected with the same velocity. If one is projected at an angle of $30^{o}$ and the other at $60^{o}$ to the horizontal, the ratio of maximum heights reached, is

1 : 3

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2 : 1

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3 : 1

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1 : 4

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Solution

The correct option is A $1 : 3$

For maximum height
$H = \frac{u^{2} {s i n}^{2} α}{2 g}$
$H_{1} = \frac{u^{2} {s i n}^{2} 30^{o}}{2 g}; H_{2} = \frac{u^{2} {s i n}^{2} 60^{o}}{2 g}$

$H_{1} : H_{2} = {s i n}^{2} 30^{o} : {s i n}^{2} 60^{o} = \frac{1 / 4}{3 / 4} = 1 : 3$

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Two projectiles are projected with the same velocity. If one is projected at an angle of 30o and the other at 60o to the horizontal, the ratio of maximum heights reached, is

The correct option is A 1:3 For maximum heightH=u2sin2α2gH1=u2sin230o2g;H2=u2sin260o2g H1:H2=sin230o:sin260o=1/43/4=1:3

Two projectiles are projected with the same velocity. If one is projected at an angle of $30^{o}$ and the other at $60^{o}$ to the horizontal, the ratio of maximum heights reached, is

The correct option is A $1 : 3$

For maximum height
$H = \frac{u^{2} {s i n}^{2} α}{2 g}$
$H_{1} = \frac{u^{2} {s i n}^{2} 30^{o}}{2 g}; H_{2} = \frac{u^{2} {s i n}^{2} 60^{o}}{2 g}$

$H_{1} : H_{2} = {s i n}^{2} 30^{o} : {s i n}^{2} 60^{o} = \frac{1 / 4}{3 / 4} = 1 : 3$