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 4.80 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.00 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 5.10 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 1.40 m/s2. How far is Powers above the ground when the helicopter crashes into the ground?

Explanation / Answer

during acceleration for 10 sec,

Applying d = v0 t + a t^2 / 2

d1 = (0 x 10) + (4.80 x 10^2 / 2 ) = 240 m

v1 = v0 + a t = 0 + (4.80 x 10) = 48 m/s

after that in free fall :

a = -9.81 m/s^2

at maximum height, velocity will be zero.

vF^2 - vi^2 =2 a d

0^2 - 48^2 = 2(-9.81)(d2)

d2 = 117.4 m

Maximum height = d1 + d2 = 357.4 m ..........Ans

time taken by copter to crash after its engine fails,

- 240 = 48t - 9.81 t^2 /2

9.81 t^2 /2 - 48 t - 240 = 0

t = 13.43 sec

height of powers after 5.10 sec :

d - 240 = 48t - 9.81(5.10^2 / 2)

d = 357.4

at this time, speed = 0

time remaining for powers, t = 13.43 - 5.10 = 8.33 s

y - 357.4 = 0 + (1.40 x 8.33^2 /2 )

y = 309 m ..........Ans

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