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 121 m/s and are 98.0 m above the ground, what distance before the target should they drop the package? 27.1 m 777 m 541 m 305 m
Question
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 121 m/s and are 98.0 m above the ground, what distance before the target should they drop the package? 27.1 m 777 m 541 m 305 m
Solution
To solve this problem, we need to find out how long it will take for the package to hit the ground after it's dropped. We can use the equation of motion:
d = 0.5gt^2
where: d = distance (98.0 m) g = acceleration due to gravity (approximately 9.8 m/s^2) t = time
Rearranging the equation to solve for time gives us:
t = sqrt((2d)/g)
Substituting the given values:
t = sqrt((2*98.0)/9.8) = 4.48 seconds
This is the time it will take for the package to hit the ground.
Now, we need to find out how far the plane will travel in this time. We can use the equation:
d = vt
where: d = distance v = velocity (121 m/s) t = time (4.48 seconds)
Substituting the given values:
d = 121 * 4.48 = 541.68 m
So, the plane should drop the package approximately 542 m before the target. Therefore, the closest answer is 541 m.
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