An Alaskan rescue plane traveling 44 m/s
drops a package of emergency rations from
a height of 147 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
h = 0.5*g*T^2 = 147.
T =
d = Vx*T.
d = 44*T.
To find the point where the package strikes the ground relative to the point directly below where it was released, we can use the equations of motion.
We know the initial velocity (u) of the package is 0 m/s since it was simply dropped from a height. The acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts downward). The height (h) from which the package is dropped is 147 m.
We can use the equation of motion: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
For the package to strike the ground, the displacement (s) would be the height from which it was dropped, which is -147 m (negative because it is measured downward). Since the initial velocity is 0 m/s, the equation becomes:
-147 = (1/2)(-9.8)t^2
Simplifying the equation:
-147 = -4.9t^2
Dividing both sides by -4.9:
t^2 = 30
Taking the square root of both sides:
t ≈ √30
t ≈ 5.48 seconds
Therefore, the package takes approximately 5.48 seconds to reach the ground.
To find the distance the package strikes the ground relative to the point directly below where it was released, we can use the equation s = ut + (1/2)at^2 again.
Since the initial velocity (u) is 0 m/s, the equation becomes:
s = 0 + (1/2)(-9.8)(5.48)^2
Simplifying the equation:
s = -4.9 * (5.48)^2
s ≈ -4.9 * 30
s ≈ -147
Therefore, the package strikes the ground at a distance of -147 m relative to the point directly below where it was released.
Since displacement is measured downward, the package strikes the ground 147 m below the point directly below where it was released.