Question

A body is launched from the earth's surface an angle $\alpha ={30}^o$ to the horizontal at a speed $v_0=\sqrt{\frac{1.5GM}{R}}$. Neglecting air resistance and earth's rotation, find the height to which the body will rise is given as $h=(\frac{\sqrt{x}}{2}+1)R$. Find $x$

Answer 1

$\text{Velocity along direction perpendicular to radial direction}$ $=v_0\cos \alpha$

$\text{At the maximum height, only velocity along the direction}$ $\text{perpendicular to the radial direction exists. Let it be}u$.

$\text{From conservation of angular momentum,}$

$m\left(v_0\cos \alpha \right)R=mur$

$⟹u=v_0\cos \alpha \frac{R}{r}$

$\text{From conservation of energy, we have}$

$\frac{1}{2}m{v}_0^2-\frac{GMm}{R}=\frac{1}{2}m{u}^2-\frac{GMm}{r}$

$⟹\frac{1}{2}m\left(\frac{3}{2}\frac{GM}{R}\right)-\frac{GMm}{R}=\frac{1}{2}m\left(\frac{3}{2}\frac{GM}{R}\right)\left(\frac{3}{4}\right)\left(\frac{R^2}{r^2}\right)-\frac{GMm}{r}$

$⟹4r^2-16Rr+9R^2=0$

$⟹r=(2+\frac{\sqrt{7}}{2})R$

Hence height , $h=r-R=(1+\frac{\sqrt{7}}{2})R$