Question 1177418: A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 7.0 m/s^2.
Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 12.0 s, 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 the 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 7.0 s after leaving the
helicopter, and then he has a constant downward acceleration with magnitude 2.0 m/s^2.
How far is Powers above the ground when the helicopter crashes into the ground?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Absolutely, let's break down this action-packed physics problem step-by-step!
(a) What is the maximum height above ground reached by the helicopter?
First, we need to determine the height reached by the helicopter during the time when it is accelerating upward with Dr. Evil and Austin Powers onboard.
Given:
* Initial velocity (vi) = 0 m/s (since the helicopter starts from rest)
* Acceleration (a) = 7.0 m/s²
* Time (t) = 12.0 s
Using the equation of motion for uniformly accelerated linear motion:
```
d = vi*t + 1/2*a*t²
```
Plugging in the values:
```
d = 0*12.0 + 1/2*7.0*12.0² = 504 m
```
Now, we need to determine the distance traveled by the helicopter during its free-fall motion after the engine is shut off.
Given:
* Initial velocity (vi) = vi (the final velocity at the end of the upward acceleration phase)
* Acceleration (a) = -9.8 m/s² (acceleration due to gravity)
* Final velocity (vf) = 0 m/s (at the maximum height)
Using the equation of motion:
```
vf² = vi² + 2*a*d
```
Plugging in the values:
```
0 = vi² + 2*(-9.8)*d
```
Solving for d:
```
d = vi² / (2*9.8)
```
To find vi, we can use the equation:
```
vf = vi + a*t
```
Plugging in the values:
```
vf = 0 + 7.0*12.0 = 84 m/s
```
Substituting this value back into the equation for d:
```
d = 84² / (2*9.8) ≈ 360 m
```
Therefore, the maximum height above ground reached by the helicopter is 504 m (initial height) + 360 m (free-fall height) = 864 m.
(b) How far is Powers above the ground when the helicopter crashes into the ground?
First, we need to determine the time it takes for the helicopter to crash after the engine is shut off.
Given:
* Initial velocity (vi) = 84 m/s (the final velocity at the end of the upward acceleration phase)
* Acceleration (a) = -9.8 m/s² (acceleration due to gravity)
* Distance (d) = -504 m (negative since it's falling downwards)
Using the equation of motion:
```
d = vi*t + 1/2*a*t²
```
Plugging in the values:
```
-504 = 84*t + 1/2*(-9.8)*t²
```
Solving for t using the quadratic formula:
```
t ≈ 19.5 s
```
Now, we need to determine how far Powers travels during the 7.0 seconds before he deploys his jet pack.
Given:
* Initial velocity (vi) = 84 m/s (the final velocity at the end of the upward acceleration phase)
* Acceleration (a) = -9.8 m/s² (acceleration due to gravity)
* Time (t) = 7.0 s
Using the equation of motion:
```
d = vi*t + 1/2*a*t²
```
Plugging in the values:
```
d = 84*7.0 + 1/2*(-9.8)*7.0² ≈ 352.1 m
```
Therefore, Powers is 504 m (initial height) + 352.1 m (free-fall height) = 856.1 m above the ground when he deploys his jet pack.
Next, we need to determine how far Powers travels during the remaining time until the helicopter crashes.
Given:
* Initial velocity (vi) = 0 m/s (since he starts from rest after deploying the jet pack)
* Acceleration (a) = -2.0 m/s² (acceleration due to jet pack)
* Time (t) = 19.5 s (total time for helicopter to crash) - 7.0 s (time before jet pack) = 12.5 s
Using the equation of motion:
```
d = vi*t + 1/2*a*t²
```
Plugging in the values:
```
d = 0*12.5 + 1/2*(-2.0)*12.5² ≈ -156.3 m
```
Therefore, Powers travels 156.3 m downwards during this time.
Finally, we can determine how far Powers is above the ground when the helicopter crashes:
* Distance above ground = 856.1 m (height when jet pack is deployed) - 156.3 m (distance traveled downwards) = 699.8 m
Therefore, Powers is approximately **699.8 meters** above the ground when the helicopter crashes.
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