Question

Two particles are projected vertically upwards from the surface of the earth with velocities $v_1=\sqrt{\frac{2gR}{3}}$ and $v_2=\sqrt{\frac{4gR}{3}}$ respectively. If the maximum heights attained by the two particles are $h_1$ and $h_2$ respectively, then calculate the ratio $\frac{h_2}{h_1}$ .

Answer 1

Answer: 4

Loss of $KE=$ Gain in $PE$
$\Rightarrow \frac{1}{2}m{v}^2=mgh\left(\frac{R}{h+R}\right)$
CASE I.
$\Rightarrow \frac{1}{2}m\left(\frac{2gR}{3}\right)={\left(mgh\right)}_1\left(\frac{R}{h_1+R}\right)$ ...(i)
$\Rightarrow \frac{1}{2}m\left(\frac{4gR}{3}\right)={\left(mgh\right)}_2\left(\frac{R}{h_2+R}\right)$ ...(ii)
Solving (i)
$\Rightarrow \frac{1}{3}=\frac{h_1}{h_1+R}$
$\Rightarrow h_1=\frac{R}{2}$ ...(iii)
Solving (ii)
$\Rightarrow \frac{2}{3}=\frac{h_2}{h_2+R}$ ...(iv)
$\Rightarrow h_2=2R$
From (iii) and (iv)
$\Rightarrow \frac{h_1}{h_2}=\frac{R/2}{2R}$
$\therefore \frac{h_1}{h_2}=\frac{1}{4}=0.25$