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Toranc3

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## Homework Statement

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s^2. 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 the ground reached by the helicopter?

B Powers deploys a jet pack strapped on his back 7.0 seconds after leaving the helicopter, and then he has constant downward acceleration with magnitude 2.0m/s^(2). How far is powers above the ground when the helicopter crashes into the ground?

## Homework Equations

y=yo+vo*t + 1/2*a*t^(2)

Vy=Vo +a*t

vy^(2)=vo^(2) + 2a(y-yo)

## The Attempt at a Solution

A)

y=yo+vo*t + 1/2*a*t^(2)

y=1/2*5.0m/s^(2)*100s^(2)=250m

vy=vo+a*t

vy=5.0m/s^(2) * 10=50m/s

vy^(2)=vo^(2) + 2a(y-yo)

0=(50m/s)^(2)+2(-9.81m/s^(2))*y

y=127.4209m + 250m =380mFor part B though I was a bit confused with a part.

This is what I have

y=yo+vo*t + 1/2*a*t^(2)

y=250m + 50m/s*7s + 1/2*(-9.81m/s^(2)) * 49s^(2)

y=359.655m

Here is where I am confused

In the problem it says this:

After the two men struggle for 10.0s , Powers shuts off the engine and steps out of the helicopter

at 10 seconds the helicopter is 250m above the ground and at that point powers leaves the helicopter right? How did his height from the ground increase to 359.655m? Did shutting off the engine take some time?

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