# Homework Help: Dive-bomber projectile motion problem

1. Sep 17, 2008

### jessedevin

1. The problem statement, all variables and given/known data

A dive-bomber has a velocity of 230 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.30 km. Find the angle θ.

2. Relevant equations
xf=xi+1/2(vxi +vxf)*t
xf=xi+vxi*t+1/2a*t2
vxf2= vxi2+2*a(xf-xi)
vf = vi + a*t
vxi= vi*cos(θ)
vyi= vi*sin(θ)
h= vi2*sin2 (θ)/(2g), g= gravity constant 9.8 m/s2, h= max height
R= vi2*sin(2θ)/g, R=max horizontal range

3. The attempt at a solution

I tried to do this by creating a triangle and using angle laws to find θ, where I did θ=tan-1(2.15/3.30), but thats wrong. I just cant visualize how to start this problem. If someone can just give me a hint on starting the problem, then maybe i can attempt to get a solution. I need help asap! Thanks!

Last edited: Sep 18, 2008
2. Sep 18, 2008

### jessedevin

No one can help me out? Not even how to start?

3. Sep 18, 2008

### Kurdt

Staff Emeritus
You know the horizontal and vertical components of initial velocity depend on the angle of the plane. Now you have to use the info about the magnitude of the displacement to find the angle. You should come up with two equations (one for the horizontal and one for the vertical component of motion) that will have two unknowns. You can then eliminate one of the unknowns and solve for the angle.

4. Sep 18, 2008

### jessedevin

The horizontal component of velocity is vxi= vi*cos(θ) and the vertical component of velocity is vyi= vi*sin(θ), so vxi=230cos(θ) and vyi=230sin(θ). Im up with you until that part. But how would you relate that with the magnitude of displacement. I am really confused. Please elaborate please!

5. Sep 18, 2008

### jessedevin

If we create a big triangle from the point of release, the hypotenuse I think will be 3.30 km. The altitude, or y, of when the bomb is released is 2.15 km, and judging by the question, θ is the angle at which the bomb hits the target. This is the way im visualizing it. Is this the way you are looking at it? And how would the components of velocity help find θ?

6. Sep 18, 2008

### Kurdt

Staff Emeritus
The only time the angle appears is when the question states the plane makes the angle below the horizontal. You're then asked to find the angle (i.e. the angle the plae is at wrt the horizontal.

Ok so you have the velocities. Now what equations describe how far the bomb travels in each direction?

7. Sep 18, 2008

### jessedevin

Well, we have the equations xf=xi+vxi*t+1/2a*t2 to describe the horizontal postion of the bomb and yf=yi+vyi*t+1/2a*t2 to describe the vertical position of the bomb. This thing we need to find out time t when the bomb hits the target, but how do we do that?

Last edited: Sep 18, 2008
8. Sep 18, 2008

### jessedevin

The x postion eqn will have no acceleration, and we can substitute vxi with vi*cos(θ), but we still have a problem with t. For the y position, we can substitute the acceleration with the gravity constant and vyi with vi*sin(θ), but again is there a substitution we can make for t? I was thinking of the formula vf = vi + a*t, where we can say the final velocity is 0, so we can say t= -vi/a, but for the x postion, a=0, so Im still stuck. Am I even in the right direction to solve this problem?

9. Sep 18, 2008

### Kurdt

Staff Emeritus
Sticking with the distance equations, if we knew the horizontal or x distance travelled we would have two equations with two unknowns (time and angle), we could then eliminate one of the unknowns and find the other. Can you think of how we can work out the horizontal distance travelled? (hint: think about the displacement given)

10. Sep 18, 2008

### LowlyPion

What again is the displacement? And what is displacement?

They say Total displacement, so that means displacement in x and y taken together. Any ancient Greeks that come to mind?

If you know 2 sides of a triangle how do you find the third?

11. Sep 18, 2008

### jessedevin

We should use the pythagrean thrm, in which:
x= $$\sqrt{}$$3.32-2.152 = 2.50 km
So now we know x= 2.5 km, y=2.15 km, and the displacement = 3.3. But how does this eliminate one of the varibles, in this case t, for the position equations?

12. Sep 18, 2008

### Kurdt

Staff Emeritus
For the x axis manipulate the equation so t is the subject. Substitute t in the y axis equation and then you have a single equation with only one unknown, the angle. Solve for the angle.

13. Sep 18, 2008

### LowlyPion

Well for one thing you now know that Vx*t = 2.5km.

14. Sep 18, 2008

### jessedevin

I tried that, also keeping in mind that i changed the units from m to km for all distances, and I got:
xf=xi + vxi*t +1/2*a*t^2, but the 1/2*a*t^2=0 because a=0, so I got t= xf/vxi, which equals t= xf/(vi*cos(θ)). When I put in numbers, I get t= 10.9/cos(θ).
Then I use the yf=yi+ vyi*t +1/2*a*t^2, and substitute in for t.
I get:
0= 2.15 +.23sin(θ)(10.9/cos(θ))-1/2(9.8)(10.9/cos(θ))^2
But I still cannot solve for θ.

Do you have any other suggestions? Am I doing this right?

15. Sep 18, 2008

### Kurdt

Staff Emeritus
Ok you're now gonna have to do some trig substitutions to see if you can get a quadratic in tan theta. Then solve the quadratic. Careful with your signs.

16. Sep 18, 2008

### jessedevin

I tried it on my ti-89 also and it couldnt even solve it! What to do ? Any advice from anyone else? LowlyPion.... HELP! lol

17. Sep 18, 2008

### Kurdt

Staff Emeritus
I'm afraid that's what you have to do unfortunately. Its a standard solution that you're obviously first encountering. Have a look at these trig identities and try and get all tans in you equation.

http://en.wikipedia.org/wiki/Trigonometric_identity

It will then be a quadratic in tan and quadratics are easy to solve using the quadratic formula.

18. Sep 18, 2008

### LowlyPion

Careful. Better leave things in meters. Or change your g = 9.8 to km.

19. Sep 18, 2008

### jessedevin

Okay i did that and still cannot get an answer? Do you have any tips on a different way to solve this problem? Its due tommorrow so if you could really help I would appreciate it!

20. Sep 18, 2008

### Kurdt

Staff Emeritus
Have you tried making the equation all tan thetas and solving the quadratic? If you're looking for someone to give you the answer then this is not the forum. I gave you pretty much everything you need to work with. If you're having trouble with a final answer post your working so far and we can see if you've made any mistakes.

21. Sep 18, 2008

### jessedevin

Ill show you exactly what i did; Im not asking you to give me the answer just a way to get the right answer!:

x2+y2= displacement2
x2=3.32-2.152= 6.27 km
x= 2.50 km --> 2500 m.
vxi= vi*cos(θ) = 230*cos(θ)
vyi=vi*sin(θ)= 230*sin(θ)
xf= xi +vi*cos(θ) +1/2*a*t2 , 1/2*a*t2=0, because a=0
2500= 0 + 230*cos(θ)*t
t=10.9/cos(θ)
yf= yi + vi*cos(θ) - 1/2*g*t2
0= 2150 + 230*sin(θ)(10.9/cos(θ)) -.5(9.8)(10.9/cos(θ))2

Do you have any way to solve for θ? Did I do any calculations wrong? Please Help!!!

22. Sep 18, 2008

### Kurdt

Staff Emeritus
You're fine up until then. Now refer to my posts #15 and #17. If there is anything you don't understand in them, just ask and I'll clarify. Can you get tan thetas in you equations using the trig identities page I linked you to? What is tan theta defined as for example?

23. Sep 18, 2008

### jessedevin

I got θ = 91.5 or θ=6.72, is that what you got?

24. Sep 18, 2008

### jessedevin

Thanks for the help! I finalllllllllllllllllly got it! lol Thanks for the help

25. Sep 18, 2008

### LowlyPion

Congratulations then.

Cheers.