A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 50.0m above the glacier at a speed of 100m/s . How far short of the target should it drop the package?
h = 0.5g*t^2 = 50 m.
4.9t^2 = 50
t^2 = 10.2
Tf = 3.19 s. = Fall time.
d = V*Tf = 100m/s * 3.19s = 319 m.
t= √ 2y/g = √2(10m)/9.8m/s^2 = √2(10)/9.8 = 3.19 s
d= 3.19s(100) = 319.4 = 319m
To find how far short of the target the plane should drop the package, we need to calculate the horizontal distance the package will travel while the plane is at a height of 50.0m above the glacier.
Since the package will fall vertically, the time it takes to reach the ground is determined by the equations of motion. We can use the equation:
h = ut + (1/2)gt^2
Where,
h = initial height of the package
u = initial velocity of the package (zero since it is dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Substituting the values, we have:
50.0 = 0*(t) + (1/2)*(-9.8)*(t^2)
Simplifying the equation, we get:
-4.9t^2 + 50 = 0
Now, let's solve the quadratic equation to find the value of t.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac))/2a, where a = -4.9, b = 0, and c = 50.
t = (-0 ± √(0^2 - 4*(-4.9)*50))/(2*(-4.9))
t = (√(0 - (-980)))/(9.8)
t = √(980)/9.8
t = 10/9.8
t ≈ 1.02 seconds
So, it takes approximately 1.02 seconds for the package to reach the ground.
Now, we can calculate the horizontal distance traveled by the package during this time:
Horizontal distance = velocity * time
Horizontal distance = 100 * 1.02
Horizontal distance ≈ 102 meters
Therefore, the plane should drop the package approximately 102 meters short of the target.
To find out how far short of the target the plane should drop the package, we need to consider the horizontal distance covered by the plane during the time it takes for the package to fall to the ground.
First, let's calculate the time it takes for the package to fall vertically from the plane's altitude of 50.0 m to the ground. We can use the equation for free fall:
h = (1/2) * g * t^2
Where:
h = vertical displacement (altitude) = 50.0 m
g = acceleration due to gravity = 9.8 m/s^2
t = time
Rearranging the equation to solve for time (t):
t^2 = (2h) / g
t = sqrt((2 * 50.0) / 9.8)
t ≈ 3.19 seconds (rounded to two decimal places)
Now, we will calculate the horizontal distance covered by the plane during this time. The horizontal distance is given by the formula:
distance = speed * time
distance = 100 m/s * 3.19 s
distance ≈ 319 meters
Therefore, the plane should drop the package 319 meters short of the target to compensate for the horizontal distance covered by the plane while the package falls.