A canister is dropped from a helicopter 500m above the ground

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Discussion Overview

The discussion revolves around the physics problem of a canister dropped from a helicopter 500 meters above the ground, specifically focusing on whether it will burst upon impact given its designed impact velocity of 100 m/s. The scope includes mathematical reasoning and problem-solving approaches related to kinematics.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Technical explanation

Main Points Raised

  • The original poster (OP) calculates the time of fall and the impact velocity using the equation s(t)=500-4.9t^2, arriving at a velocity of approximately -99.127 m/s, suggesting the canister can withstand the impact.
  • Another participant corrects the OP's velocity calculation to 98.9947 m/s, asserting that the canister does not burst either way.
  • Some participants suggest using a more precise value for gravitational acceleration (9.81 m/s²) instead of 9.8 m/s² to refine the calculations.
  • There are claims that calculus is unnecessary for solving the problem, as the formulas for distance and velocity can be used directly.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of calculus for solving the problem and the accuracy of the initial calculations. There is no consensus on the correct approach or final outcome regarding whether the canister will burst.

Contextual Notes

Some calculations depend on the precision of gravitational acceleration and the interpretation of the problem statement. The discussion includes varying methods of solving the problem, with no resolution on the best approach.

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A canister is dropped from a helicopter 500m above the ground...

A canister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100m/s. Will it burst or not?

I did the problem, but the back of the book does not agree with me. Where did I go wrong?

s(t)=500-4.9t^2 => s(t)=0 => t=10.1015 secs
I then differentiate to find velocity, which is v(t)=-9.8t, I plug in t from previous => v(10.1015)= -9.8(10.1015)= -99.127m/s. So I said it is able to withstand the impact, but the book says "No." Did I go wrong someplace? Thanks!

*EDIT*
Haha, this is embarrassing, I read the question wrong, I read "Will it burst or not?" as will it survive, so I guess that was the correct method.
 
Last edited:
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gsphysics said:
A canister is dropped from a helicopter 500m above the ground...

A canister is dropped from a helicopter 500m above the ground. Its parachute does not open, but the canister has been designed to withstand an impact velocity of 100m/s. Will it burst or not?

I did the problem, but the back of the book does not agree with me. Where did I go wrong?

s(t)=500-4.9t^2 => s(t)=0 => t=10.1015 secs
I then differentiate to find velocity, which is v(t)=-9.8t, I plug in t from previous => v(10.1015)= -9.8(10.1015)= -99.127m/s. So I said it is able to withstand the impact, but the book says "No." Did I go wrong someplace? Thanks!

*EDIT*
Haha, this is embarrassing, I read the question wrong, I read "Will it burst or not?" as will it survive, so I guess that was the correct method.
I read this twice. The question has been answered but with a slight error

I get 98.9947m/s

Its does not burst either way
 
Since it is so close, try the calculation with one more decimal place. 9.81 rather than 9.8.
 
mathman said:
Since it is so close, try the calculation with one more decimal place. 9.81 rather than 9.8.
I started with the initial conditions and used the OPs approximation but solved without differentiating (it's been a while). I did say it was slight.
 
Calculus is completely unnecessary. You have the formulas for distance and velocity.
 
mathman said:
Calculus is completely unnecessary. You have the formulas for distance and velocity.
Yes, solved without using it.
 

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