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 153 m/s and are 126 m above the ground, what distance before the target should they drop the package?
h = 0.5g*t^2.
126 = 4.9t^2, t = 5.07 s. to reach gnd.
d = Vx*t = 153 * 5.07 = 775.85 m.
Well, it sounds like these pilots have quite a "groundbreaking" task on their hands! But fear not, I'm here to assist with a touch of comedy.
To determine the distance before the target where they should drop the package, we're going to rely on our good friend Physics. Specifically, we can employ the kinematic equation:
d = v0 * t + (1/2) * a * t^2
Now, since the package is being dropped vertically, its initial velocity (v0) will be the same as the plane's velocity (153 m/s) in this case. The acceleration (a) due to gravity is a constant -9.8 m/s^2, and we can assume the time (t) it takes to reach the target is the same as it takes for the package to free-fall from 126 m.
So, plugging in the values we have:
126 = 153 * t + (1/2) * (-9.8) * t^2
Now it's time for some mathematical magic! Solving this equation will give us the time it takes for the package to fall.
To determine the distance before the target where the package should be dropped, we need to find the horizontal distance traveled by the plane before the package lands on the target.
Let's use the equation d = v*t, where:
- d is the horizontal distance traveled,
- v is the velocity of the plane, and
- t is the time it takes for the package to fall.
To find the time it takes for the package to fall, we'll use the formula for the vertical distance traveled by a falling object:
h = (1/2) * g * t^2,
where:
- h is the vertical distance,
- g is the acceleration due to gravity, approximately 9.8 m/s^2, and
- t is the time it takes for the package to fall.
We can rearrange this equation to solve for t:
t^2 = (2 * h) / g,
t = sqrt((2 * h) / g).
Now let's substitute the given values into the formulas:
h = 126 m,
v = 153 m/s,
g = 9.8 m/s^2.
First, calculate the time it takes for the package to fall:
t = sqrt((2 * 126) / 9.8) ≈ 5.05 s.
Now we can find the horizontal distance traveled:
d = v * t = 153 m/s * 5.05 s ≈ 772.65 m.
Therefore, the plane should drop the package approximately 772.65 meters before the target.
To determine the distance before the target where the package should be dropped, we need to consider the horizontal distance covered by the package while it falls vertically.
First, let's calculate the time it takes for the package to fall from a height of 126 m. We can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (change in the vertical position) = -126 m (as it goes downwards)
u = initial velocity = 0 m/s (as it starts from rest)
a = acceleration due to gravity = -9.8 m/s^2 (as it is in the downward direction)
t = time
By plugging in the values and rearranging the equation, we can solve for t:
-126 = 0 * t + (1/2)(-9.8)t^2
-126 = -4.9t^2
t^2 = 126 / 4.9
t^2 ≈ 25.7143
t ≈ √25.7143
t ≈ 5.07 seconds
Now that we have the time it takes for the package to fall, we can calculate the horizontal distance it covers:
distance = velocity * time
distance = 153 m/s * 5.07 s
distance ≈ 775.71 m
Therefore, the plane should drop the package approximately 775.71 meters before the target to ensure it lands perfectly.