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 5.60 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 9.60 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 effects of air resistance. What is the maximum height above ground reached by the helicopter? Powers deploys a jet pack strapped on his back 7.00 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 1.80 m/s2. How far is Powers above the ground when the helicopter crashes into the ground?

Explanation / Answer

Before the engine shuts off the velocity gained by the helicopter v = at = 5.6*9.6 = 53.76 m/s.

The height before the engine shuts off, h = 0.5*5.6*9.6^2 = 258.05 m

a)

So the maximum height = h0 + v0^2/2g = 258.05 + 53.76^2/(2*9.8) = 405.51 m.

b)

Time taken by helicopter to reach the ground = v0/g + sqrt(2*hmax/g) = 53.76/9.81+sqrt(2*405.51/9.81) = 14.57 sec.

in 7s after leaving the helicopter distance travelled = 53.76*7-0.5*9.8*7^2 = 136.22 m upward. The velocity at this moment = 53.76-9.8*7 = 14.84 m/s downward.

Now in next 7.57sec the acceleration is constant = 1.8 so distance travelled = 14.84*7.57+0.5*1.8*7.57^2 = 163.91 m.

So they will be at 258.05+136.22-163.91 = 230.36 m above the ground.

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