An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims above.
a. How many seconds before the plane is directly overhead should the box be dropped?
b. What is the horizontal distance between the plane and the victims when the box is dropped?
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To solve both of these questions, we need to consider the horizontal and vertical components of the airplane's motion. Let's break it down step by step:
a. How many seconds before the plane is directly overhead should the box be dropped?
We can start by converting the given speed of the airplane from km/h to m/s:
125 km/h = (125 * 1000m) / (3600s) ≈ 34.72 m/s
The time it takes for the package to reach the victims will be the same as the time it takes for the plane to cover the horizontal distance between them. So, we need to find the time it takes for the plane to reach the vertical position directly above the victims.
The vertical distance the plane has to travel is 1001 m. Since we are looking for the time in seconds, we can use the formula:
time = distance / velocity
time = 1001 m / 34.72 m/s ≈ 28.8 seconds
Therefore, the box should be dropped approximately 28.8 seconds before the plane is directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
To find the horizontal distance, we can multiply the horizontal speed of the plane by the time it takes to reach the vertical position above the victims.
distance = speed * time
distance = 34.72 m/s * 28.8 s ≈ 1000.2 meters
Therefore, the horizontal distance between the plane and the victims when the box is dropped is approximately 1000.2 meters.