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Standard XII

Physics

V-t to a-t Graph

A body is fir...

Question

A body is fired with a velocity of magnitude $\sqrt{g R} < V < \sqrt{2 g R}$ at an angle of $30^{o}$ with the radius vector of the earth. If at the highest point, the speed of the body is $V / 4$ , the maximum height attained by the body is equal to
where $R =$ radius of earth. (Use the appropriate assumptions).

\frac{V^{2}}{8 g}

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R

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\sqrt{2} R

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none of these

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Solution

The correct option is C $R$
Conservation of angular momentum of the body about O yields
$(m V s i n 30^{0}) R = m V^{'} (R + h)$
$\frac{V}{2} R = \frac{V}{4} (R + h) [A s V^{'} = \frac{V}{4}]$
$\frac{V}{2} R = \frac{V R}{4} + \frac{V h}{4}$
$\frac{R}{4} = \frac{h}{4}$
$∴ h = R$

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A body is fired with a velocity of magnitude √gR<V<√2gR at an angle of 30o with the radius vector of the earth. If at the highest point, the speed of the body is V/4, the maximum height attained by the body is equal towhere R= radius of earth. (Use the appropriate assumptions).

The correct option is C RConservation of angular momentum of the body about O yields (mVsin300)R=mV′(R+h)V2R=V4(R+h)[AsV′=V4]V2R=VR4+Vh4R4=h4∴h=R

The correct option is C $R$
Conservation of angular momentum of the body about O yields
$(m V s i n 30^{0}) R = m V^{'} (R + h)$
$\frac{V}{2} R = \frac{V}{4} (R + h) [A s V^{'} = \frac{V}{4}]$
$\frac{V}{2} R = \frac{V R}{4} + \frac{V h}{4}$
$\frac{R}{4} = \frac{h}{4}$
$∴ h = R$