To solve the problem of finding the distance AB where a body dropped from an airplane strikes the ground, we can follow these steps:
### Step 1: Convert the velocity of the airplane from km/h to m/s
The velocity of the airplane is given as \(600 \, \text{km/h}\). We need to convert this to meters per second (m/s) using the conversion factor \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\).
\[
v_x = 600 \, \text{km/h} \times \frac{5}{18} \, \text{m/s} = \frac{600 \times 5}{18} \, \text{m/s} = \frac{3000}{18} \, \text{m/s} \approx 166.67 \, \text{m/s}
\]
### Step 2: Calculate the time of flight
The body is dropped from a height of \(1960 \, \text{m}\). We can use the equation of motion in the vertical direction to find the time of flight. The vertical motion can be described by the equation:
\[
s_y = u_y t + \frac{1}{2} a_y t^2
\]
Where:
- \(s_y = 1960 \, \text{m}\) (the height from which the body is dropped)
- \(u_y = 0 \, \text{m/s}\) (initial vertical velocity)
- \(a_y = -9.81 \, \text{m/s}^2\) (acceleration due to gravity)
Substituting the values into the equation:
\[
1960 = 0 \cdot t + \frac{1}{2} \cdot (-9.81) \cdot t^2
\]
This simplifies to:
\[
1960 = -4.905 t^2
\]
Rearranging gives:
\[
t^2 = \frac{1960}{4.905} \approx 399.18
\]
Taking the square root:
\[
t \approx \sqrt{399.18} \approx 20 \, \text{s}
\]
### Step 3: Calculate the horizontal distance AB
Now that we have the time of flight, we can calculate the horizontal distance \(AB\) using the formula:
\[
d = v_x \cdot t
\]
Where:
- \(v_x \approx 166.67 \, \text{m/s}\) (horizontal velocity)
- \(t \approx 20 \, \text{s}\) (time of flight)
Substituting the values:
\[
d = 166.67 \, \text{m/s} \cdot 20 \, \text{s} \approx 3333.33 \, \text{m}
\]
### Final Answer
The distance \(AB\) is approximately \(3333.33 \, \text{m}\).
---
