The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the bank is __________ cm. (Take $g=10$ m/s²)

<p>The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank. </p> <p>The velocity of the boat in still water is $27 \, \text{km/h}$. Since the boat is moving downstream and the river flows at $9 \, \text{km/h}$, the actual speed of the boat relative to the bank becomes:</p> <p>$ v_{\text{boat}} = 27 + 9 = 36 \, \text{km/h} $</p> <p>Convert this speed from km/h to m/s by using the conversion factor $1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}$:</p> <p>$ v_{\text{boat}} = 36 \times \frac{5}{18} = 10 \, \text{m/s} $</p> <p>This is also the horizontal velocity of the ball as seen by the observer on the bank.</p> <p>Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is $10 \, \text{m/s}$. The time to reach the maximum height $t_{\text{up}}$ is given by:</p> <p>$ t_{\text{up}} = \frac{v_{\text{initial}}}{g} = \frac{10}{10} = 1 \, \text{s} $</p> <p>The total time of flight $t_{\text{total}}$ is twice the time to reach the maximum height:</p> <p>$ t_{\text{total}} = 2 \times 1 = 2 \, \text{s} $</p> <p>The horizontal range $R$ of the ball is the horizontal velocity multiplied by the total time of flight:</p> <p>$ R = v_{\text{horizontal}} \times t_{\text{total}} = 10 \times 2 = 20 \, \text{m} $</p> <p>Finally, convert this range into centimeters:</p> <p>$ R = 20 \times 100 = 2000 \, \text{cm} $</p> <p>Thus, the range of the ball as observed by an observer at rest on the bank is $2000 \, \text{cm}$.</p>