Helicopter height and vertical speed?

[bbbl67]

Homework Statement

This is not my homework question, I was asked to help with it, but I've been out of the engineering field for many years now. Here's the question:

Starting from rest, a 2500 kg helicopter accelerates straight up at a constant 1.7 m/s2. What is the helicopter's height at the moment its blades are providing an upward force of 29 kN? The helicopter can be modeled as a 2.6-m -diameter sphere, and air resistance is not negligible.

Homework Equations

Gravity:
F_g = m g

Drag:
C_d = F_d/(1/2 ρ u^2 A) |
F_d | drag force
C_d | drag coefficient
ρ | mass density
u | characteristic speed
A | frontal area

I looked up the drag coefficient (https://is.gd/qpe40K) of a sphere, and found two figures:
C_d = 0.1 (laminar)
C_d = 0.45 (turbulent)

So, I chose the turbulent figure (0.45).

The Attempt at a Solution

We know that the two forces will add up to 29 kN at a certain point, so:
F_g + F_d = 29 kN
F_d = 29 kN - F_g
= 29 kN - m g
= 29 kN - 2500 kg * 9.8 m/s^2
= 4.5 kN

The standard density of air at sea level is:
ρ = 1.204 kg/m^3

The frontal area of a sphere is:
A = pi * r^2
= pi * (2.6 m / 2)^2
= 5.309 m^2

So we have everything except the vertical speed of the helicopter, which we can find by rearranging the drag coefficient formula, and we get:
C_d = F_d/(1/2 ρ u^2 A)
u^2 = F_d/(1/2 ρ C_d A)
u = sqrt(F_d/(1/2 ρ C_d A))
= sqrt(4.5 kN/(0.5 * 1.204 kg/m^3 * 0.45 * 5.309 m^2))
= 56 m/s

Now to find the final height of the helicopter, we have to use and rearrange the acceleration-velocity-distance formula:

v_f^2 = v_i^2 + 2 a d |
v_f | final speed
a | acceleration
v_i | initial speed
d | distance

v_i = 0
v_f = u = 56 m/s
a = 1.7 m/s^2
d = h = ?
h = (v_f^2 - v_i^2)/ 2a
= (56 m/s)^2 / 2 (1.7 m/s^2)

Therefore,
h = 883 m

Were my procedures okay? Any critiques?

 

[mfb]

F_g + F_d = 29 kN

That would make the helicopter keep its speed.
Don't forget its acceleration.

 

[bbbl67]

That would make the helicopter keep its speed.
Don't forget its acceleration.

Well, the acceleration was used in the height calculation.

 

[mfb]

That is not the only step where you need it. I quoted the part where you forgot to consider it.