Solving Flight & Velocity Problem in Cutnell & Johnson Physics

[mmiller39]

This is a problem from Cutnell and Johnson Physics

I have set up my formula so that eventually I am solving a quadtratic equation, I am unsure if my calculations are correct, any input would be appreciated.

THE QUESTION:

An airplane is flying with a velocity of 240 m/s at an angle of 30.0° with the horizontal. 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 ?

WHAT WE KNOW:

Vo: 240 m/s
y = 2.4 km
Theta: 30 degrees
Beta : The unknown angle

MY CALCULATIONS:

Vo Sin 30 = Voy = -120 m/s (because it is in a downward direction, I think)

y = Voy(t) + 1/2ay(t^2)

-2,400 = -120(t) + 1/2(-9.80)(t^2)

In quadratic Form:

-4.9t^2 - 120(t) + 2,400 = 0

t = -120 +/- Sqrt (120^2 - 4(-4.9 x 2400))/2 x -4.9

THIS IS WHERE I GET CONFUSED

I am unsure if my signs are correct or if I am solving the formula correctly. I will go ahead and show the rest of my calculations but they are incorrect. (or could have been from the beginning)


t = 13.04 s

Vo cos 30 = Vox = 207 m/s

x = Vox(t) = 2699 m

Beta = tan^-1 y/x = tan^-1 2,400/2,699 = 41.64 degrees. <-----

*this answer is wrong, I am unsure where I made my error(s). Any help is greatly appreciated.

 

[dmoravec]

what you computed at the very end is the angle if you were to take the spot where the plane let the flare go and where it hit. Because of gravity the flare does not fall in a straight line. You need to compute the instantaneous velocities in both the x and y direction (you already have the x velocity). The y one will just be $$v_y(t) = v_{oy}+g*t$$ making sure that the negative signs are correctly placed. Then you can compute the angle as beta=tan^-1 v(y)/v(x) where v(y) and v(x) are the velocities at impact.

 

[mmiller39]

what you computed at the very end is the angle if you were to take the spot where the plane let the flare go and where it hit. Because of gravity the flare does not fall in a straight line. You need to compute the instantaneous velocities in both the x and y direction (you already have the x velocity). The y one will just be $$v_y(t) = v_{oy}+g*t$$ making sure that the negative signs are correctly placed. Then you can compute the angle as beta=tan^-1 v(y)/v(x) where v(y) and v(x) are the velocities at impact.

Thanks...this is a big help although it confused me as well.

You say I already have the X velocity, I am not sure what that is, Secondly here is my calculations of what you informed me on, am I placing the signs in the correct places?

-120 + -9.80(13.04) = Vy(t)

-247.79 = Vy(t)
Vy = -19.00 <-----this does not seem right to me.

 

[dmoravec]

well the x velocity never changes (there is no force in the x direction) so your inital x velicty [Vo cos 30 = Vox = 207 m/s] is also your final x velocity.

Your work is correct until the last step. I meant Vy(t) as a function of t, not mutiplied by t. Therefore your y velocity at t=13.04 seconds would be -247.79.