Ahelicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. determine the acceleration of the helicopter and its initial velocity.

Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a =  the acceleration of descent., m/s².
Let u =  initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u,  v, and a are measured as positive upward.

Then
 u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a                  (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600       (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

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