# Projectile with fixed angle

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.

## Answers and Replies

pinsky
Could you write the mathematical proof? Perhaps something could be concluded from it.
Try using $$tags please. squash Sure. I should have included this originally. target position = w arrow position = c coordinate system = arrow fired at x=0, y=0 in +x, +y direction, cliff at x=20 height y=28.5 At point of intersection, wx = cx and wy = cy We have wx = cx 20 = tv0cos[tex]\theta$$
and so tv0 = 20/cos$$\theta$$ ----------

And wy = cy
28.5 - 4.9t2 = tv0sin$$\theta$$ - 4.9t2
28.5 = tv0sin$$\theta$$ --------------

Substituting  into  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!