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

An observer on the river bank perceives the ball's velocity as the vector sum of the boat's velocity and the ball's upward velocity. The ball's horizontal velocity is identical to the boat's velocity, which is \( 9 \, {km/h} = 2.5 \, {m/s} \).The duration for the ball to ascend to its peak altitude is determined by:\[t = \frac{v_{{up}}}{g} = \frac{10}{10} = 1 \, {second}\]The horizontal distance covered is calculated by multiplying the horizontal velocity by the total time of flight:\[{Range} = 2.5 \times 1 = 100 \, {cm}\]Therefore, the accurate result is \( 100 \, {cm} \).