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 . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 13.0 , 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. Part A) What is the maximum height above ground reached by the helicopter? ANSWER = 820 m (solved it already) Part B) Powers deploys a jet pack strapped on his back 7.0 after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 . 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*10^2 = 250m The velocity of the helicopter is now 5*10 = 50m/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 = = 250 + 50*7 - 1/2*9.8*7^2 = 360m and the velocity = vy0 - g*t = 50 - 9.8*7 = -18.6m/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 = 360 -18.6*t - 4.9*t^2 so 4.9*t^2 + 18.6*t -360 = 0 so t = 6.88s Now in 6.88s we find Powers height y = y0 + vy0*t - 1/2*2*t^2 y = 360 -18.6*6.88 -1/2*2*6.88^2 = 185m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.