A helicopter is flying horizontally at 7.5 m/s and an altitude of 17 m when a package of emergency medical supplies is ejected horizontally backward with a speed of 13 m/s relative to the helicopter. Ignoring air resistance, what is the horizontal distance between the package and the helicopter when the package hits the ground?
You can figure the time in air from the 17 m. YOu know the initial horizontal speed( -4.5m/s) find then the distance it goes horizontally. Find the distance the helicopter went in the same time.
To find the horizontal distance between the package and the helicopter when the package hits the ground, we need to analyze the motion of the package horizontally.
Let's assume that the time taken by the package to reach the ground is t. During this time, the helicopter also moves horizontally at a constant velocity of 7.5 m/s.
The horizontal distance covered by the package (d) is given by the formula:
d = v*t
where v is the horizontal velocity of the package relative to the ground.
The horizontal velocity of the package relative to the ground can be found by adding the horizontal velocity of the helicopter (which is 7.5 m/s) to the horizontal velocity of the package relative to the helicopter (which is -13 m/s, since it is ejected horizontally backward).
So, the horizontal velocity of the package relative to the ground is:
v = 7.5 m/s - 13 m/s = -5.5 m/s
Since the package is moving backward, the velocity is negative.
Now, we can find the time taken for the package to hit the ground by using the formula for vertical motion:
h = (1/2)*g*t^2
where h is the initial altitude (17 m), g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time taken.
Substituting the given values into the equation, we get:
17 = (1/2)*9.8*t^2
Simplifying the equation, we have:
t^2 = (17*2) / 9.8
t^2 = 34 / 9.8
t^2 ≈ 3.47
Taking the square root of both sides, we find:
t ≈ √3.47
t ≈ 1.86 s
Now, we can substitute the value of t into the formula for the horizontal distance covered by the package:
d = v*t
d = -5.5 m/s * 1.86 s
d ≈ -10.21 m
Since the distance traveled cannot be negative, the magnitude of the horizontal distance is:
|d| ≈ 10.21 m
Therefore, the horizontal distance between the package and the helicopter when the package hits the ground is approximately 10.21 meters.