1. Velocity of the boat relative to ground

The boat moves downstream, so its speed relative to the bank is

$v_{bg} = 27 + 9 = 36\ \text{km/h}$

Convert to m/s:

$36\ \text{km/h} = 36\times \frac{5}{18} = 10\ \text{m/s}$

So, the boat (and the man) move horizontally with speed

$u_x = 10\ \text{m/s}$

2. Throwing the ball vertically upward

The man throws the ball vertically upward with speed $10\ \text{m/s}$ relative to the boat.

Since the boat is moving horizontally at $10\ \text{m/s}$, the ball’s initial velocity relative to the ground is:

• Horizontal component: $u_x = 10\ \text{m/s}$
• Vertical component: $u_y = 10\ \text{m/s}$

3. Time of flight of the ball

For vertical projectile motion,

$T = \frac{2u_y}{g} = \frac{2\times 10}{10} = 2\ \text{s}$

4. Horizontal range as seen from the bank

The observer on the bank sees the ball move horizontally with constant speed $10\ \text{m/s}$ for $2$ s.

Hence range,

$R = u_x T = 10\times 2 = 20\ \text{m}$

Convert to cm:

$20\ \text{m} = 2000\ \text{cm}$

5. Final answer

$\boxed{2000}$