A helicopter is flying horizontally at 8.2 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?
To find the horizontal distance between the package and the helicopter when the package hits the ground, we need to determine the time it takes for the package to hit the ground.
Since the package is ejected horizontally backward with a speed of 13 m/s relative to the helicopter, its initial horizontal velocity is -13 m/s (negative due to the backward direction).
We can use the formula, h = ut + (1/2)gt^2, where h is the vertical displacement, u is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
Given:
Initial vertical velocity, u = 0 m/s (since the package is ejected horizontally)
Vertical displacement, h = -17 m (negative because the package falls downward)
Acceleration due to gravity, g = -9.8 m/s^2 (negative because it acts downward)
-17 = (1/2)(-9.8)t^2 [Substituting the given values into the formula]
-17 = -4.9t^2 [Simplifying the equation]
t^2 = 17/4.9
t ≈ 1.92 seconds [Taking the square root of both sides]
Now, we can use this time to calculate the horizontal distance traveled by the package.
Horizontal distance = horizontal velocity × time
Horizontal velocity = 8.2 m/s (given in the problem)
Horizontal distance = (8.2 m/s) × (1.92 seconds)
Horizontal distance ≈ 15.74 meters
Therefore, the horizontal distance between the package and the helicopter when the package hits the ground is approximately 15.74 meters.
To find the horizontal distance between the package and the helicopter when the package hits the ground, we need to determine how long it takes for the package to reach the ground.
Let's assume the time it takes for the package to hit the ground is 't'.
We know that the initial vertical velocity of the package is 0, as it was ejected horizontally backward from the helicopter.
Using the kinematic equation for vertical motion, we can find the time 't':
y = y0 + v0yt + (1/2)at²
Since the package is falling vertically, the initial vertical position (y0) is 17 m, the initial vertical velocity (v0y) is 0, and the vertical acceleration (a) is the acceleration due to gravity, approximately 9.8 m/s².
Plugging in these values:
0 = 17 + 0t + (1/2)(9.8)(t²)
Rearranging the equation gives:
4.9t² = 17
Now, we can solve for 't':
t² = 17 / 4.9
t ≈ √(17 / 4.9)
t ≈ 1.49 seconds
So, it takes approximately 1.49 seconds for the package to hit the ground.
Now, to find the horizontal distance between the package and the helicopter when the package hits the ground, we can use the horizontal velocity and the time 't' calculated above.
The horizontal distance is given by:
x = v0x * t
Since the helicopter is moving horizontally at a velocity of 8.2 m/s, the horizontal velocity of the package (v0x) is also 8.2 m/s.
Plugging in these values:
x = 8.2 * 1.49
x ≈ 12.17 meters
Therefore, the horizontal distance between the package and the helicopter when the package hits the ground is approximately 12.17 meters.