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

Question

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2 . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 15.0 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.

1) What is the maximum height above ground reached by the helicopter?

2) 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 1.0 m/s2 . How far is Powers above the ground when the helicopter crashes into the ground?

Explanation / Answer

The helicopter reaches a height of 1/2*a*t^2 = 1/2*5*15^2 = 562.5m

The velocity of the helicopter is now 5*15 = 75m/s upward which is now the initial velocity of Powers and the helicopter when the engine is shut off


After 7 s both are together at an altitude =y = y0 + vy0*t - 1/2*g*t^2 =

= 562.5 + 75*7 - 1/2*9.8*7^2 = 847.4m

and the velocity = vy0 - g*t = 75 - 9.8*7 = 6.4m/s

We need the time for the helicopter to strike the ground after Powers leaves

y = y0 + vy0*t - 1/2*g*t^2
or = 847.4 +6.4*t - 4.9*t^2

so 4.9*t^2 - 6.4*t -847.4 = 0 so t = 13.81 s

Now in 13.81s we find Powers height

y = y0 + vy0*t - 1/2*2*t^2

y = 847.4 +6.4*13.81 -1/2*2*13.81^2 =


=745.06m

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