Finding the Angle of Release for a Dive-Bomber in 2D Motion

[missbinky]

Homework Statement

A dive-bomber has a velocity of 280 m/s at an angle of [tex]\vartheta[/tex] 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 [tex]\vartheta[/tex].

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!

 

[cristo]

Try drawing a diagram to set the problem up.

 

[missbinky]

I get a major brain-fart everytime I try to draw it out. I don't 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 [tex]\vartheta[/tex].
3) Find [tex]\vartheta[/tex] by equating y(x=xf = -h, and using the trigonometric relation.

1 = 1 + tan^2 [tex]\vartheta[/tex]
cos^2 [tex]\vartheta[/tex]

 

[learningphysics]

Write the equation for vertical displacement in terms of time.

Write the equation for horizontal displacement in terms of time.