# Motion in 2d question

1. Sep 20, 2007

### missbinky

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

A dive-bomber has a velocity of 280 m/s at an angle of $$\vartheta$$ below the horizontal. When the altitude of the aircraft is 2.15km 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 3.25km. Find the angle $$\vartheta$$.

3. The attempt at a solution

I am really lost with this question, so if some one could give me some clues that would be wonderful. I also don't really know what the question looks like, so its hard to know where to start.

Thanks for the help!

2. Sep 20, 2007

### cristo

Staff Emeritus
Try drawing a diagram to set the problem up.

3. Sep 20, 2007

### missbinky

I get a major brain-fart everytime I try to draw it out. I dont know to even draw it out.

The teacher also gave the hints.

1) Find the horizontal distance x
2) Obtain the expression for y as a function of x for all angles $$\vartheta$$.
3) Find $$\vartheta$$ by equating y(x=xf = -h, and using the trigonometric relation.

1 = 1 + tan^2 $$\vartheta$$
cos^2 $$\vartheta$$

Last edited: Sep 20, 2007
4. Sep 21, 2007

### learningphysics

Write the equation for vertical displacement in terms of time.

Write the equation for horizontal displacement in terms of time.