Equation for Projectile Motion of Flare Fired from Accelerating Helicopter

[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

 

[gneill]

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

 

[McKeavey]

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

 

[gneill]

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

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.

 

[McKeavey]

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

 

[gneill]

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

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.

 

[McKeavey]

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

 

[gneill]

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

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.

 

[McKeavey]

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

 

[gneill]

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.

 

[McKeavey]

Um
d = v1 + (.5)at^2
V1² =v2²-2ad?
:S

 

[McKeavey]

V1y = 2at² - dy/t
d = ½(a)t²
?

 

[gneill]

d = ½(a)t²
?

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?

 

[McKeavey]

Vheli = (a)heli(30s)?

 

[gneill]

Vheli = (a)heli(30s)?

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

V_y = a*t

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

 

[McKeavey]

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

 

[PAllen]

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

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