Kinematic problem: VERY complicated!

  • Thread starter mclame22
  • Start date
  • #1
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Homework Statement


A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s². Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance.

a) What is the maximum height above ground reached by the helicopter?
b) Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0m/s². How far is Powers above the ground when the helicopter crashes into the ground?


Homework Equations


Vf = Vi + at
Vf² = Vi² + 2ad
d = Vit + 1/2at²


The Attempt at a Solution


I already did part a. to this problem and the maximum height reached by the helicopter is 380m (correct answer). The second part is driving me batty; I've been going at it for over 2 hours. This is what I've done so far, but I just can't get the right answer.

HELICOPTER:
Velocity of helicopter when Powers jumps out:
Vf = Vi + at
Vf = (0m/s) + (+5.0m/s²)(10.0s) = +50.0m/s

Time taken for helicopter to reach max height:
Vf = Vi + at
t = (Vf - Vi)/a = (0m/s - 50.0m/s)/(-9.8m/s²) = 5.10s

Time taken for helicopter to crash from max height:
Vf = Vi + at
d = Vit + 1/2at²
(-380m) = (0m/s)t + 1/2(-9.8m/s²)t²
t² = 380m/19.6m/s²
t = 4.40s

Total time from when Powers jumps out until crash:
t total = 5.10s + 4.40s = 9.50s

POWERS:
Velocity of Powers right before he uses jet pack:
Vf = Vi + at
Vf = (+50m/s) + (-9.8m/s²)(7.0s) = -18.6m/s

Distance Powers has fallen before using jet pack:
Vf² = Vi² + 2ad
(-18.6m/s)² = (+50m/s)² + 2(-9.8m/s²)d
d = -109.9m

Time until crash right as Powers uses jet pack:
Total time from when Powers jumps out until crash = 9.50s
Time taken to use jet pack = 7.0s
Time until crash right as jet pack is used = 9.50s - 7.0s = 2.50s

Distance traveled by Powers from using jet pack until crash:
d = Vit + 1/2at²
d = (-18.6m/s)(2.50s) + 1/2(-2.0m/s²)(2.50s)²
d = -210.25m

Total distance Powers has fallen when helicopter crashes = -109.9m - 210.25m = -320.15m
Height of Powers when helicopter crashes = 380m - 320.15m = 59.85m = 60m

The online assignment is not accepting this as an answer. What am I doing wrong? Was I supposed to take Power's initial velocity as he jumped from the helicopter as being 0m/s?
 
Last edited:

Answers and Replies

  • #2
557
1
I think you're not doing it correctly.
The helicopter rises with a constant acceleration to reach max height, for a given time.
They give you both a and t. Period. Why do you mess it with gravity g ?
 
  • #3
Delphi51
Homework Helper
3,407
10
(-380m) = (0m/s)t + 1/2(-9.8m/s²)t²
t² = 380m/19.6m/s²
Looks like it should be t² = 380/4.9

Distance Powers has fallen before using jet pack:
Vf² = Vi² + 2ad
(-18.6m/s)² = (+50m/s)² + 2(-9.8m/s²)d
d = -109.9m
Check the calc; I get +110. Due to the high initial velocity Powers goes up, not down. Check with
d = di + Vi*t + .5*a*t² = 250 + 50*7 - 4.9*7² = 360 m above ground.
 
  • #4
13
0
Quinzio: I did take the +5.0m/s^2 acceleration into consideration for when the engine was on.. Then the helicopter was under the influence of gravity alone when the engine cut off.

Thanks Delphi51, I did do a few calculations wrong. I ended up getting an answer of 180m which was correct. Thank you for your help!
 

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