A helicopter 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

Answer:

The acceleration is 0.0134 m/s² downward.

The initial velocity is 4.0 m/s upward.