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

\frac{1}{4}

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\frac{1}{3}

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\frac{1}{5}

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\frac{1}{9}

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Solution

The correct option is A $\frac{1}{3}$
$\frac{H_{1}}{H_{2}} = \frac{{sin}^{2} θ_{1}}{{sin}^{2} θ_{2}} = \frac{{sin}^{2} 30^{o}}{{sin}^{2} 60^{o}} = \frac{1}{3}$

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

The correct option is A 13H1H2=sin2θ1sin2θ2=sin230osin260o=13

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

The correct option is A $\frac{1}{3}$
$\frac{H_{1}}{H_{2}} = \frac{{sin}^{2} θ_{1}}{{sin}^{2} θ_{2}} = \frac{{sin}^{2} 30^{o}}{{sin}^{2} 60^{o}} = \frac{1}{3}$