1. Set up the motion equations

Take upward as positive and ground as $y=0$.

• Particle 1 is dropped from height $h$ with initial velocity $0$: $y_1=h-\frac{1}{2}gt^2$

• Particle 2 is thrown upward from ground with speed $\sqrt{2gh}$: $y_2=\sqrt{2gh}\,t-\frac{1}{2}gt^2$

They collide when $y_1=y_2$.

2. Find collision time

Set the positions equal: $h-\frac{1}{2}gt^2=\sqrt{2gh}\,t-\frac{1}{2}gt^2$ $h=\sqrt{2gh}\,t$ $t_c=\frac{h}{\sqrt{2gh}}=\sqrt{\frac{h}{2g}}$

3. Find collision height

Substitute $t_c$ into either equation: $y_c=h-\frac{1}{2}g\left(\frac{h}{2g}\right)=h-\frac{h}{4}=\frac{3h}{4}$

So the collision occurs at height $\frac{3h}{4}$ above the ground.

4. Velocity of each particle at collision

For particle 1: $v_1=0-gt_c=-g\sqrt{\frac{h}{2g}}=-\sqrt{\frac{gh}{2}}$

For particle 2: $v_2=\sqrt{2gh}-gt_c=\sqrt{2gh}-\sqrt{\frac{gh}{2}}=\sqrt{\frac{gh}{2}}$

Thus, just before collision, the two equal masses have equal and opposite velocities.

5. Velocity just after completely inelastic collision

Since masses are equal and stick together, conserve momentum: $m v_1 + m v_2 = (2m)V$ $m\left(-\sqrt{\frac{gh}{2}}\right)+m\left(\sqrt{\frac{gh}{2}}\right)=0$ So, $V=0$

Hence the combined mass comes to rest momentarily at height $\frac{3h}{4}$.

6. Time for combined mass to reach ground

After collision, it starts from rest at height $\frac{3h}{4}$ and falls under gravity.

Using $s=\frac{1}{2}gt^2$ with $s=\frac{3h}{4}$: $\frac{3h}{4}=\frac{1}{2}gt^2$ $t^2=\frac{3h}{2g}$ $t=\sqrt{\frac{3h}{2g}}=\sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$

So, in units of $\sqrt{\frac{h}{g}}$, the answer is $\boxed{\sqrt{\frac{3}{2}}}$

7. Option check

• A: $\sqrt{\frac{1}{2}}$ ❌
• B: $\frac{1}{2}$ ❌
• C: $\sqrt{\frac{3}{2}}$ ✅
• D: $\sqrt{\frac{3}{4}}$ ❌