Question

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.

Answer 1

The acceleration of the helicopter is 0.133 m/s^2 and its initial velocity is -39.9 m/s (downward direction).

Step-by-step explanation:

To determine the acceleration and initial velocity of the helicopter, we can use the equations of motion.

Given:

Height (h) = 600 m

Time (t) = 5.00 minutes = 300 seconds

Final velocity (v) = 0 m/s

Using the equation h = ut + 0.5at^2, we can substitute the given values and solve for the acceleration (a). Rearranging the equation, we get:

a = (2h)/(t^2) = (2 * 600)/(300^2) = 0.133 m/s^2

Next, using the equation v = u + at (where u is the initial velocity), we can substitute the known values and solve for u:

0 = u + (0.133 * 300)

u = -39.9 m/s

Therefore, the acceleration of the helicopter is 0.133 m/s^2 and its initial velocity is -39.9 m/s (negative sign indicates downward direction).

Answer 2

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.