Free Fall Motion - Max Height Reached by Helicopter

[clope023]

[SOLVED] free fall motion

Homework Statement

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s^2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0s , Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance.

maximum height above ground reached by the helicopter?

Homework Equations

1) (vy)^2 = (v0y)^2 +2ay(y-y0)

2) vy=v0y+ayt

3) y = y0 + v0yt +1/2ayt^2

The Attempt at a Solution

1) used equation 2 to derive the velocity

v0y = (5m/s^2)(10s) = 50m/s

plugged into equation 3 with ay = -9.8m/s^2, got:

y = 0 + (50m/s)(10s) + 1/2(-9.8m/s^2)(10s)^2 = 10m wrong

2) used equation 1 with v0y = 50m/s and ay = -9.8m/s^2 and solved for y:

y = (v0y)^2/(2*ay) = (50m/s)^2/(2*-9.8m/s^2) = 127.6m wrong

3) used equation 2 with ay = 5m/s^2 with v0y = 50m/s to solve for time

t = v0y/g = (50m/s)/(9.8m/s^2) = 5.1s

then plugged in t into equation 3 to get:

y = 0 + (50m/s)(5.1s) + 1/2(-9.8m/s^2)(5.1s)^2 = 127.6m wrong again

any and all help is really appreciated, I think the main problem is I have the wrong velocity but I'm not how else to find it.

 

[Feldoh]

1) used equation 2 to derive the velocity

v0y = (5m/s^2)(10s) = 50m/s

plugged into equation 3 with ay = -9.8m/s^2, got:

y = 0 + (50m/s)(10s) + 1/2(-9.8m/s^2)(10s)^2 = 10m wrong

You got the velocity correct, however you can't assume it takes 10s for the helicopter to reach the ground, and you didn't account for how high the helicopter was when the engines were shut off.

[tex]0 = y_0 + (50m/s)x + 1/2(-9.8m/s^2)x^2[/tex] The helicopter will follow a parabolic path (well kinda) which means that however long it takes to hit the ground half of that total time and the helicopter will be at it's max point.

Or if you know some calculus you could find where the first derivative is zero.

 

[Doc Al]

You are closer than you think.

1) used equation 2 to derive the velocity

v0y = (5m/s^2)(10s) = 50m/s

Good! This is the speed of the helicopter when it begins freefall.

plugged into equation 3 with ay = -9.8m/s^2, got:

y = 0 + (50m/s)(10s) + 1/2(-9.8m/s^2)(10s)^2 = 10m wrong

Two problems using this approach:
- You don't know the time it takes to reach max height. It's not 10s!
- The initial height is not 0. It's been rising for 10s.

2) used equation 1 with v0y = 50m/s and ay = -9.8m/s^2 and solved for y:

y = (v0y)^2/(2*ay) = (50m/s)^2/(2*-9.8m/s^2) = 127.6m wrong

This is good. What's wrong is thinking that you're starting from ground level. But you're not. What is the height of the helicopter after the initial 10s?

3) used equation 2 with ay = 5m/s^2 with v0y = 50m/s to solve for time

t = v0y/g = (50m/s)/(9.8m/s^2) = 5.1s

Ah... now you've found the time. Good!

then plugged in t into equation 3 to get:

y = 0 + (50m/s)(5.1s) + 1/2(-9.8m/s^2)(5.1s)^2 = 127.6m wrong again

Again: This is how much it rises above the height it was after 10s.

any and all help is really appreciated, I think the main problem is I have the wrong velocity but I'm not how else to find it.

You're velocity is fine. Do this: Find how high it is after accelerating upwards for 10s. That's the starting point for your free fall calculations. (You already found the speed after 10s, now find the height.)

 

[clope023]

ok, I tried this:

127.6m = y0+50m/s(10s)+ 1/2(9.8m/s^2)(10s)^2

-->

127.6m -10m = y0, y0 = 117.6m

is this the correct height, I also tried it with 5.1s but that gave me 0.

 

[Doc Al]

No. Treat the problem in two parts: (a) the part where it accelerates at + 5 m/s^2 for 10 s, and (b) the part where it accelerates at -9.8 m/s^2 until reaching maximum height.

Where is the helicopter at the end of part (a)? You already found the additional height it will climb once it starts free fall (part b), just add those two heights together.

(Don't try to solve the complete problem with one equation. You can't do it, since each part has a different acceleration.)

 

[Feldoh]

[tex]y = y_0+v_0t+\frac{1}{2}at^2[/tex]

On the way up the helicopter started on the ground so [itex]y_0 = 0[/itex] and there is no initial velocity ([itex]v_0 = 0[/itex]) when the helicopter starts it's ascent.

 

[clope023]

hmm, alright so what I got from your post was to try:

a) y = 0 + 50m/s(10s) + 1/2(5m/s^s)(10s)^2 = 750m

b) y = 0 + 50m/s(5.1s) + 1/2(-9.8m/s^s)(5.1s)^2 = 127.6m

and 750m + 127.6m = 877.6m

how's that?

edit: according to Feldoh there's is no v0, so instead maybe I'll try for a

y = 1/2(5)(10squared) = 250 and 250+127.6 = 377.6m

 

[Feldoh]

Seems right.

 

[clope023]

I got the question right! (off of masteringphysics), thanks ALOT guys I was having trouble with this one, again thanks alot!