Two bodies are projected with the same velocity. One body is projected at an angle of $30^{o}$ and the other at an angle of $60^{o}$ to the horizontal. The ratio of the maximum heights reached is

3 : 1

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1 : 3

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1 : 2

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2 : 1

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Solution

The correct option is B $1 : 3$
$H_{m a x} = \frac{v^{2} {sin}^{2} θ}{2 g}$
For same velocity:
$\frac{H_{1}}{H_{2}} = \frac{{sin}^{2} θ_{1}}{{sin}^{2} θ_{2}} = \frac{{sin}^{2} 30}{{sin}^{2} 60} = \frac{1}{3}$

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Two bodies are projected with the same velocity. One body is projected at an angle of 30o and the other at an angle of 60o to the horizontal. The ratio of the maximum heights reached is

The correct option is B 1:3Hmax=v2sin2θ2gFor same velocity:H1H2=sin2θ1sin2θ2=sin230sin260=13

Two bodies are projected with the same velocity. One body is projected at an angle of $30^{o}$ and the other at an angle of $60^{o}$ to the horizontal. The ratio of the maximum heights reached is

The correct option is B $1 : 3$
$H_{m a x} = \frac{v^{2} {sin}^{2} θ}{2 g}$
For same velocity:
$\frac{H_{1}}{H_{2}} = \frac{{sin}^{2} θ_{1}}{{sin}^{2} θ_{2}} = \frac{{sin}^{2} 30}{{sin}^{2} 60} = \frac{1}{3}$