Question

Three particles are projected vertically upward from a point on the surface of the earth with velocities $\sqrt{\left(2gR/3\right)},\sqrt{\left(gR\right)},\sqrt{\left(4gR/3\right)}$ respectively, where $R$ is the radius of the earth, and $g$ is the acceleration due to gravity on the surface of the earth. The maximum heights attained are respectively $h_1,h_2,h_3$.

• A. $h_1:h_2=2:3$
• B. $h_2=R$
• C. $h_1:h_3=1:4$
• D. None of these

Answer 1

Answer: (B), (C)

The correct options are
B $h_2=R$
C $h_1:h_3=1:4$

Calculation for $h_1$
$U_i+K_i=U_f+K_f$
$-\frac{GmM}{R}+\frac{m}{2}\frac{2gR}{3}=-\frac{GmM}{h+R}+0$
$-mgR+mg\frac{R}{3}=-mg\frac{R^2}{h+R}$
$\Rightarrow \frac{2mgR}{3}=\frac{mg{R}^2}{h+R}$
$\Rightarrow \frac{2}{3}=\frac{R}{h+R}\Rightarrow 3R=2h+2R$
$h_1=\frac{R}{2}$
Calculation for $h_2$
$-\frac{GmM}{R}+\frac{1}{2}mgR=-\frac{GmM}{h+R}+0$
$-mgR+\frac{mgR}{2}=-\frac{mg{R}^2}{R+h}$
$R+h=2R$
$\Rightarrow h_2=R$
Calculation for $h_3$
$-mgR+\frac{1}{2}\frac{m4gR}{3}=-\frac{mg{R}^2}{h+R}$
$(h+R)=3R$
$h_3=2R$