Solve Dr. Evil Problem: Austin Powers & Jetpack

[kero]

Can somebody help me please to solve this problem?

helicopter carrying Dr. Evil takes off with a constant upward acceleration of Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 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 accel-eration with magnitude How far is Powers above the ground when the helicopter crashes into the ground?

 

[tiny-tim]

Welcome to PF!

Hi kero! Welcome to PF!

Can somebody help me please to solve this problem?

helicopter carrying Dr. Evil takes off with a constant upward acceleration of Secret

we can't help you …

if the acceleration is Secret! ​

Use the usual constant acceleration equations …

show us how far you get, and where you're stuck, and then we'll know how to help!

 

[tomcue]

I would approach this problem by first breaking it down into smaller parts and using known equations and principles to solve each part individually.

For part (a), we can use the equation for displacement during constant acceleration, d = vi*t + 1/2*a*t^2, where vi is initial velocity, t is time, and a is acceleration. In this case, the initial velocity is 0 m/s and the acceleration is the constant upward acceleration of the helicopter. Solving for d, we get d = 0.5*a*t^2. Plugging in the values of a = 9.8 m/s^2 (assuming Earth's gravity) and t = 10.0 s, we get a maximum height of 490 m.

For part (b), we can use a similar equation but with a negative acceleration since Powers is now in free fall. The equation becomes d = vi*t + 1/2*(-9.8)*t^2. We also know that the initial velocity is 0 m/s and the time is 7.0 s. Solving for d, we get 0 = -4.9*t^2, which gives us a distance of 171.5 m.

Finally, to calculate the distance between Powers and the helicopter when it crashes into the ground, we can use the equation d = vi*t + 1/2*a*t^2, with vi = 0 m/s, a = -9.8 m/s^2, and t = 7.0 s. This gives us a distance of 171.5 m as well.

In conclusion, the helicopter reaches a maximum height of 490 m before Powers jumps off, and when the helicopter crashes into the ground, Powers is also 171.5 m above the ground. However, it is important to note that these calculations are based on ideal conditions and do not take into account factors such as air resistance, which may affect the actual results.