QJEEM25Ph3234: JEE Main 2025 Physics - Class 11 JEEM25Ph3234 Solution: The maximum speed of a boat in still water is 27 km/h. Now this …

The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank.

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:

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

Convert this speed from km/h to m/s by using the conversion factor $1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}$ :

$v_{\text{boat}} = 36 \times \frac{5}{18} = 10 \, \text{m/s}$

This is also the horizontal velocity of the ball as seen by the observer on the bank.

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:

$t_{\text{up}} = \frac{v_{\text{initial}}}{g} = \frac{10}{10} = 1 \, \text{s}$

The total time of flight $t_{\text{total}}$ is twice the time to reach the maximum height:

$t_{\text{total}} = 2 \times 1 = 2 \, \text{s}$

The horizontal range $R$ of the ball is the horizontal velocity multiplied by the total time of flight:

$R = v_{\text{horizontal}} \times t_{\text{total}} = 10 \times 2 = 20 \, \text{m}$

Finally, convert this range into centimeters:

$R = 20 \times 100 = 2000 \, \text{cm}$

Thus, the range of the ball as observed by an observer at rest on the bank is $2000 \, \text{cm}$ .