It takes the mailbag 7.47 seconds to reach the ground
Explanation:
The given is
1. The height of a helicopter above the ground is given by h = 2.95 t³,
where h is in meters and t is in seconds
2. At t = 1.95 s, the helicopter releases a small mailbag
The helicopter is moving up and when the mailbag releases its initial
velocity is the same with the velocity of the helicopter at this moment
So the initial velocity of the mailbag is upward then it travels up until
it reaches its maximum height and then goes down
∵ h = 2.95 t³
Find the height of the helicopter at t = 1.95 seconds
→ h = 2.95(1.95)³ = 21.87 meters
→ v = [tex]\frac{dh}{dt}[/tex]
The initial velocity of mailbag [tex]v_{0}[/tex] = [tex]\frac{dh}{dt}[/tex]
→ [tex]v_{0}[/tex] = 3(2.95) t²
→ [tex]v_{0}[/tex] = 8.85 t²
The value of t at this moment is 1.95
→ [tex]v_{0}[/tex] = 8.85(1.95)² = 33.65 m/s
When the mailbag reaches its maximum height its final velocity is 0
→ v = [tex]v_{0}[/tex] + g t
→ v = 0 m/s , [tex]v_{0}[/tex] = 33.65 m/s , g = -9.8 m/s²
→ 0 = 33.65 - 9.8 t
Add 9.8 t to both sides
→ 9.8 t = 33.65
Divide both sides by 9.8
→ t = 3.434 seconds
It takes the mailbag same time to return to its initial position when it
goes down
The time that the mailbag takes to return to the same position is
→ 2 × 3.434 ≅ 6.87 seconds
Now the mailbag goes down to the ground with initial speed 33.65 m/s
from a height 21.87 meters
→ h = [tex]v_{0}[/tex] t + [tex]\frac{1}{2}[/tex] g t²
→ h = 21.87 , g = 9.8 m/s² , [tex]v_{0}[/tex] = 33.65 m/s
Substitute these values in the rule above
→ 21.87 = 33.65 t + 4.9 t²
Subtract 21.87 from both sides
→ 4.9 t² + 33.65 t - 21.87 = 0
Solve it for t
→ t = 0.598 ≅ 0.60 seconds
The total time is the sum of the two times
→ Total time = 6.87 + 0.60 = 7.47 seconds
It takes the mailbag 7.47 seconds to reach the ground
V.I note:
You can calculate the time of trip of the mailbag in one step by using
the rule h = [tex]v_{0}[/tex] t + [tex]\frac{1}{2}[/tex] g t²
where h = -21.87 meters because the height is under the point of
released and g = -9.8 m/s² because the initial velocity is upward
→ -21.87 = 33.65 t - 4.9 t²
Add 4.9 t² to both sides and subtract 33.65 t from both sides
→ 4.9 t² - 33.65 t - 21.87 = 0
Solve it for t
→ t = 7.47 seconds
Learn more:
You can learn more about equation of trajectory in brainly.com/question/2607086
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