1. Time of fall from the post

Both the ball and the bullet leave the point of collision from the same height $h=5\,\text{m}$, and their initial vertical velocities are zero because the bullet was moving horizontally.

So, time to reach the ground is

(Using $g=10\,\text{m/s}^2$.)

2. Horizontal velocities after collision

Since horizontal distance $= v_x t$ and $t=1\,\text{s}$:

• For the ball: $v_{ball}=\frac{20}{1}=20\,\text{m/s}$

• For the bullet: $v_{bullet}=\frac{100}{1}=100\,\text{m/s}$

So after collision, the ball moves horizontally with $20\,\text{m/s}$ and the bullet with $100\,\text{m/s}$.

3. Apply conservation of horizontal momentum

During the short collision, external horizontal impulse is negligible, so horizontal momentum is conserved.

Initial horizontal momentum:

$p_i = (0.01)V$

Final horizontal momentum:

$p_f = (0.2)(20) + (0.01)(100)$

$p_f = 4 + 1 = 5\,\text{kg m/s}$

Therefore,

$0.01V = 5$

$V = \frac{5}{0.01} = 500\,\text{m/s}$

4. Check options

• A: $250\,\text{m/s}$ ✗
• B: $250\sqrt{2}\,\text{m/s}$ ✗
• C: $400\,\text{m/s}$ ✗
• D: $500\,\text{m/s}$ ✓

So the correct option is D.