Help in Part b Problem 2.89 What is the maximum height above ground reached by t

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Help in Part b

Problem 2.89 What is the maximum height above ground reached by the helicopter? Express your answer to two significant figures and include the appropriate units helicopter carrying Dr. Evil takes off with a constant upward acceleration of 7.0 m/s . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 12.0s, 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 h= 860m Submit My Answers Give Up Correct Significant Figures Feedback: Your answer 864 m was either rounded differently or used a different number of significant figures than required for this part. Part B 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 2.0 m/s. How far is Powers above the ground when the helicopter crashes into the ground? Express your answer to two significant figures and include the appropriate units h166.6m Submit My Answers Give Up Incorrect; Try Again; 7 attempts remaining vide Feedba Continue

Explanation / Answer

The helicopter reaches a height of 1/2*a*t^2 = 1/2*7*12^2 = 504m

The velocity of the helicopter is now 7*12 = 84m/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 =

= 504 + 84*7 - 1/2*9.8*7^2 = 851.9m

and the velocity = vy0 - g*t = 84 - 9.8*7 = -15.4m/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 = 851.9 -15.4*t - 4.9*t^2

so 4.9*t^2 + 15.4*t -851.9 = 0 so t = 11.7s

Now in 6.88s we find Powers height

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

y = 851.9 -15.4*11.7 -1/2*2*11.7^2 =


534.83m

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