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Question

The first ball reaches to a maximum height

$\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{g}}$

$\left(using\, v^{2}=u^{2}-2 g h\right)$

The seco

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

The first ball reaches to a maximum height

$\mathrm{h}=\frac{\mathrm{u}^{2}}{2 \mathrm{g}}$

$\left(using\, v^{2}=u^{2}-2 g hight)$

The second ball attains the maximum height

$\mathrm{h}^{\prime}=\frac{\mathrm{u}^{2} \sin ^{2} .30^{\circ}}{2 \mathrm{g}}=\frac{1}{4}\left(\frac{\mathrm{u}^{2}}{2 \mathrm{g}}ight)$

$\frac{	ext { Potential energy of first ball }}{	ext { Potential energy of sec ond ball }}=\frac{\operatorname{mgh}}{\operatorname{mgh}^{\prime}}=\frac{h}{h^{\prime}}=\frac{4}{1}$

A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected with the same speed at an angle of $30^o$ with the horizontal. At the highest point, the ratio of their potential energies is

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