Question

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 river bank is _________ cm. (Take \( g = 10 \, {m/s}^2 \)).

Hint:

When dealing with projectile motion in a moving frame of reference, remember to add the horizontal velocity of the observer (in this case, the boat’s speed) to the horizontal velocity of the projectile.

Answer 1

The observer on the river bank sees the ball’s velocity as a combination of the velocity of the boat and the upward velocity of the ball. The horizontal velocity of the ball is the same as the velocity of the boat, which is \( 9 \, {km/h} = 2.5 \, {m/s} \). The time for the ball to reach its maximum height is given by: \[ t = \frac{v_{{up}}}{g} = \frac{10}{10} = 1 \, {second} \] The horizontal range is the horizontal velocity multiplied by the time of flight: \[ {Range} = 2.5 \times 1 = 100 \, {cm} \] Thus, the correct answer is \( 100 \, {cm} \).