A helicopter carrying Dr. Evt takes off with a constant upward acceleration of 7

Question

A helicopter carrying Dr. Evt takes off with a constant upward acceleration of 7.0 m/s^2. Secret agent Austin Powers jumps on just as the helicopter lifts of the ground. After the two men struggle for 14.0 s. Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after is engine is shut off and ignore the effects of air resistance. What is the maximum height above ground reached by the helicopter? Power deploys jet pack strapped on his back 7.0 s silver 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 which the helicopter crashes into the ground?

Explanation / Answer

for 14s,

a = 7 m/s^2

v0 = 0 m/s

h = u t + a t^2 /2

h1 = (0 ) + (7 x 14^2 / 2) = 686 m

v = v0 + a t = 0 + (7 x 14) = 98 m/s

after that,

it will go up until its velocity becomes zero.

a = - 9.8 m/s^2

vf^2 - vi^2 = 2 a d

0^2 - 98^2 = 2(- 9.81) (H2)

H2 = 490 m

H = H1 + H2 = 1176 m ..........Ans (A)

(B) time taken by helicopter to crash into the gorund after engine shut off

-686 = 98t - 9.8 t^2 /2

4.9 t^2 - 98t - 686 = 0

t = 25.49 sec

displacement in 7 sec,

d1 =(98 x 7) - (9.8 x 7^2 / 2) = 445.9 m

velocity after 7 sec , v = 98 - 9.8x7 = 29.4 m/s

after that .t = 25.49 - 7 = 18.49 m

d2 = (29.4 x 18.49) - (1 x 18.49^2)/2 = 372.67 m

height from ground = h1 + d1 + d2 = 1504.6 m

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