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/s²).
To solve this problem, we'll determine the range of the ball thrown upwards from the boat as observed by a stationary observer on the riverbank. Here's a step-by-step breakdown:
Determine the total velocity of the boat and the ball:
The velocity of the boat in still water is 27 km/h. Convert this to m/s: 27 km/h = \(27 \times \frac{1000}{3600} = 7.5\) m/s.
The river flows at 9 km/h, converted to m/s: 9 km/h = \(9 \times \frac{1000}{3600} = 2.5\) m/s.
The boat's velocity relative to the riverbank is the sum of its speed and the river's speed downstream: \(7.5 + 2.5 = 10\) m/s.
The man throws the ball vertically upwards with a speed of 10 m/s; however, its horizontal velocity relative to the riverbank is equal to that of the boat: 10 m/s.
Calculate the time of flight of the ball:
Using the equation for motion under gravity, where \(v = u + gt\), and here, \(v = 0\) m/s (at the top of the trajectory), \(u = 10\) m/s (upwards), and \(g = -10\) m/s² (downwards), solve for the time to reach the maximum height: \(0 = 10 - 10t \Rightarrow t = 1\) s.
The total time of flight (up and down) is double this value: \(2 \times 1 = 2\) s.
Calculate the range observed:
Since the horizontal velocity (relative to the observer on the riverbank) remains constant at 10 m/s, the range (horizontal distance covered) is: Range = Horizontal velocity × Time of flight = \(10 \times 2 = 20\) m.
Convert the range from meters to centimeters: \(20\) m = \(2000\) cm..
Therefore, the range of the ball as observed by the observer at rest on the riverbank is 2000 cm.
