An Alaskan rescue plane traveling 42 m/s drops a package of emergency rations from a height of 117 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
006 (part 2 of 3) 10.0 points
What is the horizontal component of the ve- locity just before it hits?
Answer in units of m/s
007 (part 3 of 3) 10.0 points
What is the vertical component of the velocity just before it hits? (Choose upward as the positive vertical direction)
Answer in units of m/s
To determine where 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
where s is the vertical distance traveled, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time of flight.
We need to solve for t using the equation:
117 = 0 + (1/2)(-9.8)t^2
Simplifying:
117 = -4.9t^2
Divide both sides by -4.9:
t^2 = -117 / -4.9
t^2 = 23.8776
Taking the square root of both sides:
t = √23.8776
t ≈ 4.8866 seconds (rounded to the nearest four decimal places)
Now that we know the time of flight, we can find the horizontal component of the velocity just before it hits. The horizontal component of the velocity remains constant throughout the flight.
Given the initial horizontal velocity of 42 m/s, the horizontal component of the velocity just before it hits will also be 42 m/s.
Finally, to find the vertical component of the velocity just before it hits, we can use the equation:
v = u + at
where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time of flight.
Given:
u = 0 (as it was dropped)
a = 9.8 m/s^2 (taking upward as positive)
t ≈ 4.8866 seconds
Substituting these values into the equation:
v = 0 + (9.8)(4.8866)
v ≈ 47.8148 m/s (rounded to the nearest four decimal places)
Therefore, the vertical component of the velocity just before it hits is approximately 47.8148 m/s (upward direction).
To find where the package strikes the ground relative to the point directly below where it was released, you can use the equation:
y = initial position + initial velocity * time + (1/2) * acceleration * time^2
Where:
y = vertical distance traveled
initial position = height from which the package was dropped (117 m)
initial velocity = 0 m/s (since the package was dropped, not thrown)
acceleration = acceleration due to gravity (-9.8 m/s^2)
time = time taken for the package to reach the ground (unknown)
Since the initial velocity is zero, the equation simplifies to:
y = (1/2) * acceleration * time^2
To solve for time, rearrange the equation:
time^2 = (2 * y) / acceleration
Substitute the values into the equation:
time^2 = (2 * 117 m) / (-9.8 m/s^2)
time^2 = -23.8776
Since time cannot be negative, we can discard the negative value. Taking the square root of the positive value, we get:
time = 4.8865 s
Therefore, it takes approximately 4.8865 seconds for the package to hit the ground. Now, let's move on to the next part of the question.
To find the horizontal component of the velocity just before the package hits the ground, we can use the equation:
horizontal distance = horizontal velocity * time
The horizontal component of the velocity remains constant, as there is no horizontal acceleration. Thus, the horizontal component of the velocity just before it hits is the same as the horizontal component of the initial velocity, which is given as 42 m/s.
Therefore, the horizontal component of the velocity just before it hits is 42 m/s.
Lastly, let's calculate the vertical component of the velocity just before it hits.
The vertical component of the velocity can be found using the equation for free fall motion:
final velocity = initial velocity + acceleration * time
Where:
final velocity = velocity just before it hits (unknown)
initial velocity = 0 m/s (since the package was dropped, not thrown)
acceleration = acceleration due to gravity (-9.8 m/s^2)
time = time taken for the package to reach the ground (4.8865 s)
Substitute the values into the equation:
final velocity = 0 m/s + (-9.8 m/s^2) * 4.8865 s
final velocity = -47.9527 m/s
Since the velocity is downward, we take it as negative. Therefore, the vertical component of the velocity just before it hits is approximately -47.9527 m/s.