A body is projected up from the surface of the earth with a velocity equal to ( (3/4)th ) of its escape velocity. If R be the radius of earth, the height it reaches is

Question

A body is projected up from the surface of the earth with a velocity equal to ( (3/4)th ) of its escape velocity. If R be the radius of earth, the height it reaches is (A)  (3R/10)  (B)  (9R/7)  (C)  (8R/5)  (D)  (9R/5)

Tardigrade Admin · Accepted Answer

If body is projected with velocity (v < ve), then height upto which it will rise, h=(R/(ve2)v2-1) v=(3/4) ve (given) ∴ h=(R/(ve2)(9)16 ve2-1 =(9/7) R

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

$\frac{3 R}{10}$

$\frac{9 R}{7}$

$\frac{8 R}{5}$

$\frac{9 R}{5}$

If body is projected with velocity $(v < v_{e})$ ,
then height upto which it will rise,
$h = \frac{R}{\frac{v _{e}^{2}}{v ^{2}} - 1}$
$v = \frac{3}{4} v_{e}$ (given)
$∴ h = \frac{R}{\frac{v _{e}^{2}}{\frac{9}{16} v _{e}^{2}} - 1}$
$= \frac{9}{7} R$

Q. A body is projected up from the surface of the earth with a velocity equal to (43​th) of its escape velocity. If R be the radius of earth, the height it reaches is

103R​

79R​

58R​

59R​