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 6.0 m/s^2

Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 11.0s

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 the effects of air resistance.

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

Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 1.0 m/s^2?

How far is Powers above the ground when the helicopter crashes into the ground?

Explanation / Answer

so while accelerating will get to a height and distance

so max v = 6*11 = 66 m/s

max height = 1/2 a t^2 = 0.5*6*11^2=363 m

max height when v = 0

v^2 = v0^2 + 2 a(y-y0)

0 = 66^2 - 2*9.81*(y-363)

y=585 m

height the ground when y = 0

y = y0 + v0y t + 1/2 a t^2

0 = 363 + 66*t - 0.5*9.81*t^2

t = 17.65 s

now to powers

we need to find his speed and height after 7 s

v = 66 - 9.81*7=-2.67 m/s

y = 363 + 66*7 - 0.5*9.81*7^2=584.7 m

now he turns on his jetpack for the remaining 17.65 - 7 = 10.65 s

so y = y0 + v0y t + 1/2 a t^2 = 584.7 - 2.67*10.65 - 0.5*1*10.65^2=500 m

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