A person is standing on an open car moving with a constant velocity of 30 m/s on a straight horizontal road.The men throws a ball in the vertically upward direction and it return to the person after the car has moved 240 m. The speed and angel of projection
In the previous problem if the car moving with a constant acceleration of 2 m/`s^2` , the ball will fall behind the person at a distance

A. 32m
B. 64m
C. 96m
D. 16m

Options

A. 32m
B. 64m
C. 96m
D. 16m

Answer

To solve the problem step by step, we will analyze the motion of the car and the ball separately.

### Step 1: Analyze the initial scenario
The car is moving with a constant velocity of 30 m/s, and the person throws the ball vertically upward. The car moves 240 m while the ball is in the air.

### Step 2: Calculate the time of flight of the ball
Since the car moves 240 m at a constant speed of 30 m/s, we can find the time of flight (t) of the ball using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
Thus,
\[ 240 = 30 \times t \]
Solving for \( t \):
\[ t = \frac{240}{30} = 8 \text{ seconds} \]

### Step 3: Analyze the second scenario with acceleration
Now, the car is moving with a constant acceleration of 2 m/s². The time of flight for the ball remains the same, which is 8 seconds.

### Step 4: Calculate the distance traveled by the car during the time of flight
To find out how far the car travels during the 8 seconds, we use the formula for distance under constant acceleration:
\[ \text{Distance} = v_0 t + \frac{1}{2} a t^2 \]
where \( v_0 \) is the initial velocity (30 m/s), \( a \) is the acceleration (2 m/s²), and \( t \) is the time (8 s).

Substituting the values:
\[ \text{Distance} = 30 \times 8 + \frac{1}{2} \times 2 \times (8)^2 \]
Calculating each term:
1. \( 30 \times 8 = 240 \) m
2. \( \frac{1}{2} \times 2 \times 64 = 64 \) m

Adding these distances together:
\[ \text{Total Distance} = 240 + 64 = 304 \text{ m} \]

### Step 5: Calculate the distance the ball falls behind the person
Since the ball returns to the person after 8 seconds, we need to find how far the car has moved compared to the initial position of the ball. The ball will fall behind the person by the distance the car has traveled beyond the initial 240 m.

The distance the car has moved beyond the initial 240 m is:
\[ 304 - 240 = 64 \text{ m} \]

### Conclusion
The ball will fall behind the person at a distance of **64 m**.

### Final Answer
**B. 64 m**