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Question

Physics Question on Gravitation

A body is projected up from the surface of the earth with a velocity equal to (34th)( \frac{3}{4}th ) of its escape velocity. If RR be the radius of earth, the height it reaches is

A

3R10\frac{3R}{10}

B

9R7\frac{9R}{7}

C

8R5\frac{8R}{5}

D

9R5\frac{9R}{5}

Answer

9R7\frac{9R}{7}

Explanation

Solution

If body is projected with velocity (v<ve)\left(v < v_{e}\right),
then height upto which it will rise,
h=Rve2v21h=\frac{R}{\frac{v_{e}^{2}}{v^{2}}-1}
v=34vev=\frac{3}{4} v_{e} (given)
h=Rve2916ve21\therefore h=\frac{R}{\frac{v_{e}^{2}}{\frac{9}{16} v_{e}^{2}}-1}
=97R=\frac{9}{7} R