A plane is flying horizontally with speed
169 m/s at a height 4240 m above the ground,
when a package is dropped from the plane.
The acceleration of gravity is 9.8 m/s
2
.
Neglecting air resistance, when the package
hits the ground, the plane will be
1. directly above the package.
2. behind the package.
3. ahead of the package.
What is the horizontal distance from the release point to the impact point?
Answer in units of m.
To find the horizontal distance from the release point to the impact point, we can use the equation of motion:
displacement = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial velocity is the horizontal component of the plane's speed, which remains constant throughout the motion. The horizontal component is given by the formula:
horizontal velocity = plane speed * cos(angle)
The angle in this case is 0 degrees since the plane is flying horizontally. So the horizontal velocity is:
horizontal velocity = 169 m/s * cos(0) = 169 m/s
Next, we need to find the time it takes for the package to hit the ground. We can use the equation:
final position = initial position + initial velocity * time + (1/2) * acceleration * time^2
The initial position is the height above the ground, which is 4240 m. The final position is the height at which the package hits the ground, which is 0 m. The acceleration due to gravity is 9.8 m/s^2. Plugging these values into the equation, we get:
0 = 4240 m + 0 m/s * time + (1/2) * (-9.8 m/s^2) * time^2
Simplifying the equation, we get:
4.9 m/s^2 * time^2 = 4240 m
Solving for time, we find:
time = sqrt(4240 m / 4.9 m/s^2) = 20 s
Now we can find the horizontal distance by multiplying the horizontal velocity by the time:
horizontal distance = horizontal velocity * time = 169 m/s * 20 s = 3380 m
Therefore, the horizontal distance from the release point to the impact point is 3380 meters.