Three identical balls are projected with same speed at angle 30^o, 45^o and 60^o . Their ranges

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3-2.Motion in Plane

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Three identical balls are projected with the same speed at angle $30^o, 45^o$ and $60^o$. Their ranges are $R_1 R_2$ and $R_3$ respectively. Then

A

$R_1 = R_2 = R_3$

B

$R_1 = R_3 < R_2$

C

$R_1 < R_2 < R_3$

D

$R_1 > R_2 > R_3$

Solution

$30^{\circ}+60^{\circ}=90^{\circ}$

$\therefore \mathrm{R}_{1}=\mathrm{R}_{3}$

and we get maximum range

for $\theta=45^{\circ}$

so $\mathrm{R}_{2}$ will be maximum hence $\mathrm{R}_{1}=\mathrm{R}_{3}<\mathrm{R}_{2}$

Standard 11

Physics

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