2D Kinematics, airplanes crashing wtf?

[lmbiango]

2D Kinematics, airplanes crashing wtf??

An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle theta?

There was a diagram with this, but it's pretty self explanatory.

Vxo = 240(cos(30)) = 207.8 m/s
Vyo = 240(sin(30)) = 120 m/s

Vf^2 - Vi^2 = 2a (change in X)
Vy^2 - 120^2 = 2(9.8)(2400)
Vy^2 = 247.87 m/s

(wrong!)

and theta:
the inverse tangent of (247.87/207.8)
=50.01 degrees

Also wrong.
I have no idea where I could have gone wrong. I thought I had this one!

 

[LowlyPion]

An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal, as the drawing shows. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle theta?

There was a diagram with this, but it's pretty self explanatory.

Vxo = 240(cos(30)) = 207.8 m/s
Vyo = 240(sin(30)) = 120 m/s

Vf^2 - Vi^2 = 2a (change in X)
Vy^2 - 120^2 = 2(9.8)(2400)
Vy^2 = 247.87 m/s

(wrong!)

and theta:
the inverse tangent of (247.87/207.8)
=50.01 degrees

Also wrong.
I have no idea where I could have gone wrong. I thought I had this one!

Not self explanatory enough for me. What angle is θ in the drawing.

 

[lmbiango]

sorry, the angle is between the ground and the line of sight from the plane to the target. So it is the angle between the adjacent side (the ground) and the hypotenuse of a right triangle with the opposite side being 2.4km

 

[lmbiango]

Has everyone given up on me?! My homework is due tomorrow!

 

[LowlyPion]

sorry, the angle is between the ground and the line of sight from the plane to the target. So it is the angle between the adjacent side (the ground) and the hypotenuse of a right triangle with the opposite side being 2.4km

OK I get it now.

How long does it take the flare to drop. 2.4 KM = 1/2 a t²

Solve for t. That time times your horizontal velocity (240m/s) is the distance along the ground that the target is from the plane at Flare's Away.

Now you have a simple triangle. And theta can be found simply by the tan{SUP]-1[/SUP](Height which is opposite side / distance calculated - adjacent side) = θ

Edit: Oops. I see it is climbing. That complicates it a little, but not that much.

The horizontal velocity is the V*Cos 30.

Calculating the time is a little more involved. But not much.

You want to find the height that the flare rises to before it falls.
You know the Vertical V component. The additional height is given by V²/(2*a) = Extra height.

Add that height to 2400 meters. Figure too the time to max height. Extra Height = 1/2 a t²

Now with that extra height proceed as before and figure the time to fall from max height.
Then add the time to max height. That Total time times the horizontal speed is how far along the ground the target is.

Since it was dropped at 2400 m as before in the simple calculation, then that's the height of opposite side.

tan^-1(height/distance) = θ

 

[lmbiango]

No, actually it is landing... does that change anything?

If I do what you said in the first place though I get that the angle is 2.587 degrees, which can't be right. It's my last try so I have to make sure I don't get it wrong.

In the picture:
The plane's nose is facing down and at it's tail there is an angle of 30 degrees, so that is the direction it is pointing downward. The distance from the plane to the ground is 2.4km, but that is not the distance from the flare to the ground. The flare travels in a curve (but never goes up) to the ground. Your line of sight is labeled in the picture and that makes the hypotenuse of the right triangle you are talking about, but the flare has a curved path that starts and ends at the line of sight, but has a longer distance, and the time would also be longer i think. The reason I am having so much trouble is because the path of the flare cannot make a triangle because it isn't a line.

 

[lightgrav]

yes, but the idea is similar
(you CAN just solve the quadratic for t when flare hits,
but it's often easier to see what time it (would've) had zero vertical vel.)

you need to know the time of flare at ground level,
so you can step its hor.velocity for that length of time.

 

[lmbiango]

yes, but the idea is similar
(you CAN just solve the quadratic for t when flare hits,
but it's often easier to see what time it (would've) had zero vertical vel.)

you need to know the time of flare at ground level,
so you can step its hor.velocity for that length of time.

I don't know what you mean. Did you read what I said about the path of the flare being curved? And what quadratic are you talking about? there is only t^2 in the equation he gave me above... right?

 

[lightgrav]

when you solve an equation that has 0=h+vt+½at^2 for t, you use quadratic formula,
because it's a quadratic equation.

If the plane releases when moving downward (or upward), there is a vertical v_0 .

 

[lightgrav]

you started out (original post) having computed v_yo (but it should be negative.)
v_x stays the same all through the parabola, so you can find x if you know t_at_ground .

 

[LowlyPion]

No, actually it is landing... does that change anything?

Well ... YEAH.

You originally said:

Vxo = 240(cos(30)) = 207.8 m/s
Vyo = 240(sin(30)) = 120 m/s

OK. The idea is still the same only you don't need to worry about time to max height. But you do still care about time to hit the ground. Because that gives you your distance directly.

x = Vi*t + 1/2 at²
2400 = 240*(.5)* t + (.5)*(9.8)*t²

Solve for t.
Multiply by Vx for Distance

Then tan^-1(2400/Distance) = Angle

 

[lmbiango]

ok, so just to make sure, I solved for t and it is 13.0481.
So multiply that by Vxo (that I originally calculated, right?) and you get 2711.395m
So the angle is 41.514 degrees?

 

[LowlyPion]

ok, so just to make sure, I solved for t and it is 13.0481.
So multiply that by Vxo (that I originally calculated, right?) and you get 2711.395m
So the angle is 41.514 degrees?

I haven't done your math.

But looking at the numbers roughly, I suppose they seem about right.