Question

A body is projected horizontally from the surface of the earth ( radius =R) with a velocity equal to
n times the escape velocity neglect rotational effects of the earth, the maximum height attained by
the body from the earth's surface is R/2. then n must be
[ ]
(13)
1
2) 10.6
b)
c) 10.4
d)1÷root3

13​

Answer 1

Let escape velocity be vₑ

We have $\large\sf{v_e}$ = $\sf{\sqrt{\dfrac{2GM}{R}}}$

By angular momentum conservation:

$\sf{m{(nv_e)}R=\dfrac{3}{2}mvR\implies{v=\dfrac{2}{3}nv_e}}$

Where v is the velocity at the highest point.

By energy conservation:

$\sf \dfrac{1}{2} {m}^{2} {v_e}^{2} - \dfrac{GMm}{R} = \dfrac{1}{2} {m}^{2} {v_e}^{2} \times\dfrac{4}{9} - \dfrac{2GMm}{3R} \\ \\ \\ \sf \implies \frac{15}{18} {m}^{2} {v_e}^{2} = \dfrac{GMm}{3R} \implies{nv_e} = \sqrt{ \dfrac{6GM}{5R}} \\ \\ \\ \sf \implies{n = \sqrt{ \dfrac{3}{5} }} = \sqrt{0.6}$