An Alaskan rescue plane traveling 49 m/s
drops a package of emergency rations from
a height of 191 m to a stranded party of
explorers.
The acceleration of gravity is 9.8 m/s
2
.
Where does the package strike the ground
relative to the point directly below where it
was released?
Answer in units of m.
To find the location where the package strikes the ground relative to the point directly below where it was released, we can use the equations of motion.
First, we need to find the time it takes for the package to reach the ground. We can use the equation:
h = (1/2) * g * t^2
where h is the height (191 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Rearranging this equation to solve for t:
t^2 = (2h) / g
t^2 = (2 * 191 m) / (9.8 m/s^2)
t^2 ≈ 39.1837
t ≈ √39.1837
t ≈ 6.26 s
Next, we can find the vertical distance the package travels during this time by using the equation:
d = v * t
where d is the distance and v is the initial vertical velocity (0 m/s since the package is dropped).
Substituting the values:
d = (0 m/s) * (6.26 s)
d = 0 m
Since the vertical distance is 0, we can conclude that the package strikes the ground directly below the point where it was released. Therefore, the package strikes the ground at the same point relative to where it was dropped, and the answer is 0 m.