Assuming open chamber

$\mathrm{V}_{\text {relative }}=0.5 \mathrm{m} / \mathrm{s}$

$S_{\text {relative }}=4 \mathrm{m}$

time $=\frac{4}{0.5}=8 \mathrm{s}$

Alternate

Assuming closed chamber

In the frame of chamber :

Maximum displacement of ball A from its left end is $\frac{\mathrm{u}_{\mathrm{A}}^{2}}{2 \mathrm{a}}=\frac{(0.3)^{2}}{2(2)}=0.0225 \mathrm{m}$

This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.

Hence, time taken by ball B to reach left end will be given by

$\mathrm{S}=\mathrm{u}_{\mathrm{B}} \mathrm{t}+\frac{1}{2} \mathrm{at}^{2}$

$4=(0.2)(t)+\frac{1}{2}(2)(t)^{2}$

Solving this, we get

$\mathrm{t} \approx 2 \mathrm{s}$

As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A

$\mathrm{V}_{\mathrm{B}} \cos 30=100 \sqrt{3}$

$\mathrm{V}_{\mathrm{B}}=200$

If A is observer A remains stationary therefore

$\mathrm{t}=\frac{500}{\mathrm{V}_{\mathrm{B}} \sin 30}=\frac{500}{100}=5$

$\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 45}{2 \mathrm{g}}=120$

$\Rightarrow \frac{\mathrm{u}^{2}}{4 \mathrm{g}}=120$ ….(i)

when half of kinetic energy is lost $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$

$\mathrm{H}_{2}=\frac{\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2} \sin ^{2} 30}{2 \mathrm{g}}=\frac{\mathrm{u}^{2}}{16 \mathrm{g}}$

from (i) & (ii)

$\mathrm{H}_{2}=\frac{\mathrm{H}_{1}}{4}=30 \mathrm{m}$ on 30.00