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)

Solution

Physics sir · 3 steps

The bank watcher sees the ball carry the boat's ground speed sideways, so add the river to the boat first.

1. Horizontal speed. Downstream, the river adds to the boat's own speed. Ground speed is 27 + 9 = 36 km/h = 36 × 5/18 = 10 m/s. The ball leaves the man's hand with this same horizontal velocity.
2. Time of flight. Vertically the ball goes up at 10 m/s and comes back to the same height. So T = 2u/g = (2 × 10)/10 = 2 s.

   R = vₓ × T = 10 m/s × 2 s = 20 m

3. So. The observer on the bank sees a range of 20 m = 2000 cm.

The common slip: 1500. Using only the boat's still-water speed of 27 km/h (7.5 m/s) and leaving out the river gives 15 m, which is 1500 cm.

The idea to keep. A ball thrown straight up from a moving boat or train keeps that vehicle's ground velocity. For a ground observer, find that ground velocity first, then treat the throw as a normal projectile.

Answer from the NTA final answer key. Solution by PrepWiser.