Question

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 7.0 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 11.0 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.
A. What is the maximum height above ground reached by the helicopter?
B. 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?
C. How far is Powers above the ground when the helicopter crashes into the ground?

Answer 1

Answer:

a) 725.5 m

b) 630 m

Explanation:

Given data:

acceleration of Helicopter = 7.0 m/s^2

time spent upwards by helicopter = 11.0 seconds

a) Determine the maximum height above ground reached by the helicopter

h1  = at^2 /2

   = 7 * 11^2 / 2

   = ( 7 * 121 ) / 2 =  423.5 m  

  also v = a*t  = 7 * 11 = 77 m/s

also we calculate h2

h2 = v^2 / 2g

    = (77^2) / 2 * 9.81

    = 302 m

therefore the maximum height =  302 + 423.5 = 725.5 m

b) Given that ;  power deploys a jet pack strapped on his back at 7.0 s and with a downward acceleration of ; 1.0 m/s^2

Determine distance Power reaches before helicopter crashes

s = ut + 1/2 at^2

h.gt^2 - 77t - 423.5 m = 0

h.gt^2 - 77t = 423.5

t = 17. 66 secs

Yf = 423.5 + 77 *7 - 4.9 *7

Yf = 928.2

Vf = u + at

    = 77 - 9.8*7  = 8.4 m/secs

t' = 17.66 - 7 = 10.66 secs

hence

Yf = 725.5 - 8.4 * 10.66 + 1/2 * -1 * 10.66

   = 630 meters