Question

Three particles are projected vertically upward from a point on the surface of the earth with velocities $\sqrt{2gR/3}$, $\sqrt{gR}$, $\sqrt{4gR/3}$ 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. $h_2:h_3=3:4$
• C. $h_1:h_3=1:4$
• D. $h_2=R$

Answer 1

Answer: (C), (D)

$h_2=R$

At the maximum height, kinetic energy of the particle will become zero.
For $h_1$:
By the energy conservation,
$U_i+K_i=U_f+K_f$
$\Rightarrow -\frac{GmM}{R}+\frac{1}{2}m{\left(\sqrt{2gR/3}\right)}^2=-\frac{GmM}{h_1+R}+0$

Also, $g=\frac{GM}{R^2}$

$\Rightarrow -mgR+mg\frac{R}{3}=-mg\frac{R^2}{h_1+R}$

$\Rightarrow \frac{2mgR}{3}=\frac{mg{R}^2}{h_1+R}$

$\Rightarrow \frac{2}{3}=\frac{R}{h_1+R}$

$\Rightarrow 3R=2h_1+2R$

$h_1=\frac{R}{2}$

Similarly, For $h_2$:
By the energy conservation,
$\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mgR=-\frac{GmM}{h_2+R}+0$

$\Rightarrow -mgR+\frac{mgR}{2}=-\frac{mg{R}^2}{R+h_2}$

$\Rightarrow R+h_2=2R$

$\Rightarrow h_2=R$

Similarly, for $h_3$:
By the energy conservation,
$\Rightarrow -mgR+\frac{1}{2}\frac{m4gR}{3}=-\frac{mg{R}^2}{h_3+R}+0$

$\Rightarrow (h_3+R)=3R$

$\Rightarrow h_3=2R$

Now, $h_1:h_2=\frac{R}{2}:R=1:2$
$h_2:h_3=R:2R=1:2$
$h_1:h_3=\frac{R}{2}:2R=1:4$