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Two objects are projected at an angle $\theta^{\circ}$ and $\left(90-\theta^{\circ}\right)$, to the horizontal with the same speed. The ratio of their maximum vertical heights is
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Verified Answer
The correct answer is:
$\tan ^2 \theta: 1$
We know that, maximum vertical height in projectile motion,
$$
H=\frac{u^2 \sin ^2 \phi}{2 g}
$$
[where, $\phi$ is angle of projection.]
$$
\begin{aligned}
& \Rightarrow \quad H \propto \sin ^2 \phi \\
& \Rightarrow \quad \frac{H_1}{H_2}=\frac{\sin ^2 \phi_1}{\sin ^2 \phi_2}=\frac{\sin ^2 \theta}{\sin ^2(90-\theta)} \\
& \text { [Given, } \left.\phi=\theta \text { and } \phi_2=90^{\circ}-\theta\right] \\
& =\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta \\
& \therefore \quad H_1: H_2=\tan ^2 \theta: 1 \\
&
\end{aligned}
$$
$$
H=\frac{u^2 \sin ^2 \phi}{2 g}
$$
[where, $\phi$ is angle of projection.]
$$
\begin{aligned}
& \Rightarrow \quad H \propto \sin ^2 \phi \\
& \Rightarrow \quad \frac{H_1}{H_2}=\frac{\sin ^2 \phi_1}{\sin ^2 \phi_2}=\frac{\sin ^2 \theta}{\sin ^2(90-\theta)} \\
& \text { [Given, } \left.\phi=\theta \text { and } \phi_2=90^{\circ}-\theta\right] \\
& =\frac{\sin ^2 \theta}{\cos ^2 \theta}=\tan ^2 \theta \\
& \therefore \quad H_1: H_2=\tan ^2 \theta: 1 \\
&
\end{aligned}
$$
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