Constant Acceleration and Free Fall Problem

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SUMMARY

The discussion centers on a physics problem involving a package dropped from a helicopter traveling upward at 10.0 m/s. The package takes 3.5 seconds to reach the ground, with calculations revealing the helicopter's height at 25.025 meters when the package was released and 60.025 meters above ground when the package impacts. The maximum height reached by the package after release is debated, with one participant suggesting 30.227 meters, while another notes a discrepancy using g = 9.81 m/s², yielding 29.9 meters.

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  • Familiarity with free fall and constant acceleration concepts
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1. Homework Statement
A package is dropped from a helicopter. Note that both the helicopter and package are traveling upward at 10.0m/s when the package is released. If the package takes 3.5 seconds to hit the ground, how high was the helicopter when the package was released?
If the helicopter's velocity does no change, how far above the ground will the helicopter be located when the package hits the ground?
What is the maximum height above the ground that this package reaches after it was released by the helicopter?

2. Homework Equations




3. The Attempt at a Solution
So far I've tried to answer the first two questions question and got that the helicopter was 25.025m high when the package was released and was 60.025m above the ground when the package hit the ground.

I made an x vs. t graph, a v vs. t graph, and an a vs. t graph to help with solving these as well.

So for the last question:
"What is the maximum height above the ground that this package reaches after it was released by the helicopter?"

Would the answer be 30.227m?

Thanks for your help!
 
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All correct, I think. I get slightly different answers using g = 9.81 (29.9 for the last part).
 

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