A boy is standing on an open truck. The truck is moving with acceleration `2ms^(-2)` on a horizontal road. When the speed of the truck is `10 ms^(-1)` and reaches to an pole , boy projected a ball with velocity `10ms^(-1)` in a vertically upward direction relative to himself (take `g = 10 ms^(-2)`). Neglect the height of boy and truck. The distance of the ball from the pole where ball land is

Options

A. 20 m
B. 10 m
C. 30 m
D. 40 m

Answer

To solve the problem step by step, we need to analyze the motion of the ball projected by the boy on the moving truck.

### Step 1: Understanding the Initial Conditions
- The truck is moving with an acceleration of \(2 \, \text{m/s}^2\).
- At the moment the boy throws the ball, the speed of the truck is \(10 \, \text{m/s}\).
- The boy throws the ball vertically upward with a velocity of \(10 \, \text{m/s}\).

### Step 2: Determine the Time of Flight of the Ball
The ball is projected vertically upward with an initial velocity of \(10 \, \text{m/s}\). To find the total time the ball is in the air, we need to calculate the time taken to reach the maximum height and then the time taken to fall back to the original height.

1. **Time to reach maximum height**:
- At maximum height, the final vertical velocity \(v = 0\).
- Using the equation of motion:
\[ v = u - gt \]
where \(u = 10 \, \text{m/s}\) (initial velocity), \(g = 10 \, \text{m/s}^2\) (acceleration due to gravity).
- Setting \(v = 0\):
\[ 0 = 10 - 10t \implies t = 1 \, \text{s} \]

2. **Total time of flight**:
- The time to go up is \(1 \, \text{s}\) and the time to come down is also \(1 \, \text{s}\).
- Therefore, the total time of flight \(T\) is:
\[ T = 1 + 1 = 2 \, \text{s} \]

### Step 3: Calculate the Horizontal Distance Traveled by the Ball
The horizontal velocity of the ball at the moment it is thrown is equal to the velocity of the truck, which is \(10 \, \text{m/s}\). Since there is no horizontal acceleration acting on the ball after it is thrown, it will continue to move horizontally at this constant velocity.

- **Horizontal distance traveled**:
\[ \text{Distance} = \text{Horizontal Velocity} \times \text{Total Time} \]
\[ \text{Distance} = 10 \, \text{m/s} \times 2 \, \text{s} = 20 \, \text{m} \]

### Step 4: Conclusion
The distance of the ball from the pole when it lands is \(20 \, \text{m}\).

### Final Answer
The distance of the ball from the pole where it lands is **20 meters**.

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