An Alaskan rescue plane traveling 38 m/s drops a package of emergency rations from a height of 105 m to a stranded party of explorers. The acceleration of gravity is 9.8 m/s2. Where does the package strike the ground
relative to the point directly below where it
was released?
Answer in units of m.
t = sqrt[2(105)/9.8]
x = 38t
To find out 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 that the initial velocity of the package is 0 m/s (since it is dropped), the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward), and the initial height is 105 m.
Using the equation of motion:
s = ut + 0.5 * a * t^2
where:
s = final displacement (distance)
u = initial velocity
t = time
a = acceleration
We can rearrange the equation to solve for time (t):
0 = 0 + 0.5 * -9.8 * t^2
0 = -4.9 * t^2
Solving for t, we find that t = 0 or t = 4.29 seconds (ignoring the time t = 0 since it is not relevant to this scenario).
Now that we have the time, we can calculate the final displacement (distance) using the equation:
s = ut + 0.5 * a * t^2
s = 0 + 0.5 * -9.8 * (4.29)^2
s = -0.5 * 9.8 * 18.36
s = -89.82 m
Since the displacement is negative, it means that the package strikes the ground 89.82 meters below the point directly below where it was released.
Therefore, the package strikes the ground 89.82 meters below the point directly below where it was released.