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3-2.Motion in Plane
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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
A
$1:1$
B
$2:1$
C
$3:2$
D
$4:1$
Solution
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 second ball attains the maximum height
$\mathrm{h}^{\prime}=\frac{\mathrm{u}^{2} \sin ^{2} .30^{\circ}}{2 \mathrm{g}}=\frac{1}{4}\left(\frac{\mathrm{u}^{2}}{2 \mathrm{g}}\right)$
$\frac{\text { Potential energy of first ball }}{\text { Potential energy of sec ond ball }}=\frac{\operatorname{mgh}}{\operatorname{mgh}^{\prime}}=\frac{h}{h^{\prime}}=\frac{4}{1}$
Standard 11
Physics
