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/s²).

Answer 1

Correct Answer: 2000

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:

1. 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.
2. 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.
3. 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.

Answer 2

Step 1: Given data and conversion.
Maximum speed of the boat in still water = 27 km/h
Speed of river = 9 km/h
Speed of ball thrown vertically = 10 m/s
Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

First, convert the boat’s speed (downstream) into m/s:
\[ \text{Boat speed downstream} = 27 + 9 = 36 \, \text{km/h} \] \[ 36 \, \text{km/h} = 36 \times \frac{1000}{3600} = 10 \, \text{m/s} \] Hence, the boat moves with a horizontal velocity of \( 10 \, \text{m/s} \) relative to the ground.

Step 2: Motion analysis (in observer’s frame).
The man in the boat throws the ball vertically upward with a speed of \( 10 \, \text{m/s} \) relative to the boat. Since the boat is moving horizontally with \( 10 \, \text{m/s} \), the horizontal component of the velocity of the ball (as seen by an observer on the river bank) is also \( 10 \, \text{m/s} \).

The vertical motion is independent of the horizontal motion. The time of flight for vertical motion is:
\[ T = \frac{2u}{g} = \frac{2 \times 10}{10} = 2 \, \text{s} \]

Step 3: Horizontal range as seen from the river bank.
During this time, the ball continues moving horizontally with velocity \( 10 \, \text{m/s} \). Thus, horizontal range:
\[ R = u_x \times T = 10 \times 2 = 20 \, \text{m} \] Convert to centimeters:
\[ R = 20 \times 100 = 2000 \, \text{cm} \]

Final Answer:
\[ \boxed{2000 \, \text{cm}} \]