# Homework Help: Projectile Motion HARD

1. Sep 29, 2011

### McKeavey

A helicopter at rest on the ground started to accelerate uniformely upwards. After 30s a flare was fired horizontly from the helicopter. An observer on the ground measured the time the flare reached the ground as 10 seconds. Neglect air resistance.

A) acceleration of helicopter
B) height from which the flare was fired
c) height of the helicopter at the instant the flare reaches the ground, assuming acceleration remained constant

3.2 as acceleration is wrong
490m as distance is wrong.

I have absolutely no idea what other ways there are to solve it..
Also to know, the flare will make a parabola because of the velocity the flare takes from the helicopter

2. Sep 29, 2011

### Staff: Mentor

You haven't described your attempt to solve the problem. What did you try?

3. Sep 29, 2011

### McKeavey

Im on my phone its hard to type out my three pages worth of notes lmao

4. Sep 29, 2011

### Staff: Mentor

Well, we can't help if we can't see what you're doing! Perhaps you can follow up when you have access to a "real" keyboard and screen.

5. Sep 29, 2011

### McKeavey

Well ill try to shorten it because i need the answer asap.
y = 0 + 0.5(9.8mls^2)(10s)^2
= 490m
but thats wrong

6. Sep 29, 2011

### Staff: Mentor

The equation does not represent the motion of the flair. It is launched horizontally from the helicopter, but the helicopter is moving upwards at some velocity when the flair is released. So the flair gains a vertical component of motion thereby.

You will need to combine equations for the flair motion and helicopter motion to solve this problem.

7. Sep 29, 2011

### McKeavey

Nono the equation is only for finding out the height of when the flair was released.

8. Sep 29, 2011

### Staff: Mentor

All you know is that the flair was in the air for ten seconds. It wasn't falling straight down from rest! At first it went upwards due to the initial vertical velocity imparted by the helicopter at launch time.

You can't solve directly for the height of the flair without also knowing its vertical speed at landing, or some other parameter of the motion. That's where the motion of the helicopter comes in. You need to consider the equations for both together.

9. Sep 29, 2011

### McKeavey

Ohh..
Hm.
V2heli = V1flare?
what equations do i use :S

10. Sep 29, 2011

### Staff: Mentor

Write the equations of motion for the helicopter. One for height vs time, the other for its vertical speed vs time. Both will involve the (as yet) unknown upward acceleration of the helicopter.

11. Sep 29, 2011

### McKeavey

Um
d = v1 + (.5)at^2
:S

12. Sep 29, 2011

### McKeavey

V1y = 2at² - dy/t
d = ½(a)t²
?

13. Sep 29, 2011

### Staff: Mentor

That one's right. Only why not use an appropriate variable name for the distance? How about "h" for the height of the helicopter?

Now, what's the helicopter's velocity with respect to time?

14. Sep 29, 2011

### McKeavey

Vheli = (a)heli(30s)?

15. Sep 29, 2011

### Staff: Mentor

30s is a particular time, not time in general. Letting Vy be the vertical velocity of the helicopter, then

Vy = a*t

The velocity of the helicopter when the flair is launched will be Vy = a*30s.

16. Sep 29, 2011

### McKeavey

I need another equation to sub in for the v eq :/

17. Sep 29, 2011

### PAllen

write an equation for the flair, taking account upward velocity. You will have all values for everything in terms of helicopter acceleration and constants.