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A body of mass $m$ thrown up vertically with velocity $v_1$ reaches a maximum height $h_1$ in $t_1$ seconds. Another body of mass $2 \mathrm{~m}$ is projected with a velocity $v_2$ at an angle $\theta$. The second body reaches a maximum height $h_2$ in time $t_2$ seconds. If $t_1=2 t_2$, then ratio $\left(\frac{h_1}{h_2}\right)$ is
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$4: 1$
Maximum vertical height attained by body thrown with velocity $v_1$
$$
h_1=\frac{v_1^2}{2 g}
$$
Another body of mass $2 \mathrm{~m}$ is projected with a velocity $v_2$ at an angle $\theta$.
$\therefore$ Height attained $\left(h_2\right)=\frac{v_2^2 \sin ^2 \theta}{2 g}$
$\therefore \quad \frac{h_1}{h_2}=\frac{v_1^2}{v_2^2} \sin ^2 \theta$
But $\quad t_1=2 t_2$
$\therefore \quad \frac{V_2}{g}=2\left[\frac{V_2 \sin \theta}{g}\right]$
$\Rightarrow \quad v_1=2 v_2 \sin \theta$
$\therefore \quad \frac{h_1}{h_2}=\left(\frac{2}{1}\right)^2=\frac{4}{1}$
$$
h_1=\frac{v_1^2}{2 g}
$$
Another body of mass $2 \mathrm{~m}$ is projected with a velocity $v_2$ at an angle $\theta$.
$\therefore$ Height attained $\left(h_2\right)=\frac{v_2^2 \sin ^2 \theta}{2 g}$
$\therefore \quad \frac{h_1}{h_2}=\frac{v_1^2}{v_2^2} \sin ^2 \theta$
But $\quad t_1=2 t_2$
$\therefore \quad \frac{V_2}{g}=2\left[\frac{V_2 \sin \theta}{g}\right]$
$\Rightarrow \quad v_1=2 v_2 \sin \theta$
$\therefore \quad \frac{h_1}{h_2}=\left(\frac{2}{1}\right)^2=\frac{4}{1}$
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