A plane needs to drop supplies off to victims of natural disasters around the world. The first stop is for flood victims but they must drop the box so it lands perfectly on a target. If they are flying at 72.0 m/s and are 121 m above the ground, what distance before the target should they drop the package?
491 m
14.5 m
225 m
358 m
To determine the distance before the target at which the package should be dropped, we need to consider the horizontal distance traveled by the package while it falls vertically due to gravity. Here's how you can calculate it:
1. Find the time it takes for the package to fall from the plane to the ground:
- We can use the formula h = (1/2)gt^2, where h is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is time.
- In this case, the initial vertical velocity is 0 because the package is dropped, not thrown.
- Plugging in the values, we have 121 m = (1/2)(9.8 m/s^2)(t^2).
- Rearrange the equation to solve for t: t^2 = (2 * 121 m) / 9.8 m/s^2.
- Take the square root of both sides to get t: t = √(242 m / 9.8 m/s^2).
2. Calculate the horizontal distance traveled by the package:
- The horizontal distance is given by the formula d = v * t, where d is the distance, v is the horizontal velocity, and t is the time calculated in step 1.
- In this case, the horizontal velocity v is given as 72.0 m/s.
- Plug in the values: d = 72.0 m/s * t.
3. Calculate the distance before the target at which the package should be dropped:
- To find this distance, subtract the horizontal distance calculated in step 2 from the total distance to the target, which is not provided directly but can be determined by reverse engineering the answer choices.
- By inspecting the answer choices, we can see that the only option that is possible is 491 m:
Distance to target - Horizontal distance = 491 m - 72.0 m/s * t.
Substituting the previously calculated value for t in step 1, we have 491 m - 72.0 m/s * √(242 m / 9.8 m/s^2).
Therefore, the correct answer is 491 m.