An aeroplane flying `490 m` above ground level at `100 m//s`, releases a block. How far on ground will it strike

Options

A. `0 km`
B. `1 km`
C. `2 km`
D. None

Answer

To solve the problem of how far the block will strike the ground after being released from an aeroplane flying at a height of 490 meters and a speed of 100 m/s, we can break it down into steps.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Height of the aeroplane (h) = 490 m
- Horizontal speed of the aeroplane (v_x) = 100 m/s
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Calculate the Time of Fall (t):**
The block is dropped from a height of 490 m. We can use the second equation of motion for vertical motion to find the time it takes to fall:
\[ h = ut + \frac{1}{2}gt^2 \]
Here, the initial vertical velocity (u) is 0 (since it is dropped), so the equation simplifies to:
\[ 490 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ 490 = 4.9t^2 \]
Rearranging gives:
\[ t^2 = \frac{490}{4.9} = 100 \]
Taking the square root:
\[ t = \sqrt{100} = 10 \text{ seconds} \]

3. **Calculate the Horizontal Distance (d):**
The horizontal distance traveled by the block while it is falling can be calculated using the formula:
\[ d = v_x \cdot t \]
Substituting the values:
\[ d = 100 \text{ m/s} \cdot 10 \text{ s} = 1000 \text{ m} \]

4. **Conclusion:**
The block will strike the ground at a distance of 1000 meters from the point directly below where it was released.

### Final Answer:
The block will strike the ground **1000 meters** away from the point directly below the aeroplane.

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