Projectile Trajectory: Why Does the Angle of Fire Remain Constant?

[squash]

Homework Statement

An arrow is fired from the ground at the same instant a target is dropped off a 28.5m cliff. The arrow firing device is 20m from the base of the cliff. For any initial velocity of the arrow, the angle it is fired at remains constant (at about 54.9 degrees) in order to hit the target. Why does the angle remain constant?

I have proved this mathematically but this part of the problem requires a worded answer. I have thought it is perhaps because gravity acts equally upon the arrow and the target? But I feel this alone is not enough...

Thanks for the help.

 

[pinsky]

Could you write the mathematical proof? Perhaps something could be concluded from it.
Try using [tex] tags please.

 

[squash]

Sure. I should have included this originally.

target position = w
arrow position = c
coordinate system = arrow fired at x=0, y=0 in +x, +y direction, cliff at x=20 height y=28.5

At point of intersection, w_x = c_x and w_y = c_y

We have w_x = c_x
20 = tv₀cos$$\theta$$
and so tv₀ = 20/cos$$\theta$$ ----------[1]

And w_y = c_y
28.5 - 4.9t² = tv₀sin$$\theta$$ - 4.9t²
28.5 = tv₀sin$$\theta$$ --------------[2]

Substituting [1] into [2] gives

28.5 = (20sin$$\theta$$)/cos$$\theta$$
28.5/20 = tan$$\theta$$
$$\theta$$ = 54.9⁰

And so for any initial velocity the arrow must be fired at 54.9⁰ (v₀ and t cancel out). But why is this angle constant?

 

[Lachlan1]

a change in angle changes the ratio of the vertical to horizontal velocity components of the arrow.
mathematically, for the targets to meet over a range of vector velocities, each of these component velocities, these two variables, must keep a certain relationship to the single variable, the distance dropped of the ball. changing the ratio between component velocities, will mean a change in one or both of their own relationships to the distance dropped of the ball, causing the arrow to miss the target.

 

[squash]

That makes sense to me. Thanks Lachlan!