Wikipedia:Reference desk/Archives/Mathematics/2011 July 29

= July 29 =

shooting arrows and finding angles
I have been given this problem about angles:

''Archer A is positioned at the top of a cliff face, 40 meters directly above Archer B. Both archers release arrows at the same same time.

''Archer A fires horizontally from the cliff top at 28 m/s. Archer B fires at an unknown angle.''

Both arrows strike an identical spot on the ground, at the same time.

Show that the angle between the embedded arrows will be approximately 18.5 degrees.

It has been a long time since I learned how to do this type of question back in high school, I need help refreshing my memory.

Is it something to do with gravity being 9.8 m/s squared, and creating some sort of quadratic parabola...?

I don't know where to start, not even sure if it can be solved with the information given. Don't we need to know either Archer B's speed or angle? I'm clueless, please help.

118.208.128.137 (talk) 00:13, 29 July 2011 (UTC)


 * Have you tried to search our articles, e.g. Trajectory and Projectile motion? Take a look at these articles and the let us know how you get on. — Fly by Night  ( talk )  00:56, 29 July 2011 (UTC)
 * Does archer B also fire at 28m/s? Otherwise I think there's insufficient data. Sorry, missed the "strike at the same time" part. -- Meni Rosenfeld (talk) 06:50, 29 July 2011 (UTC)


 * Let's see:
 * We know that archer A fires horizontally, so the time taken for their arrow to fall 40 meters is sqrt(80/g) - let's call this time T.
 * The vertical velocity of A's arrow at time T is gT (which is sqrt(80g)), and its horizontal velocty is a constant 28 m/s, so the angle it makes with the vertical when it strikes the ground (assuming it always points along its velocity vector) is tan-1(28/gT).
 * During time T, B's arrow goes up and comes down again. Its change in vertical velocity is gT (just the same as for A's arrow) so, by symmetry, it's initial vertical velocity is gT/2 upwards, and its final vertical velocity is gT/2 downwards.
 * Since A's arrow and B's arrow are fired at the same time and hit the ground at the same point and the same time, the horizontal velocity of B's arrow must be 28 m/s, just the same as A's arrow (so the two arrows always have equal horizontal displacements throughout their motion).
 * So when B's arrow hits the ground, it makes an angle tan-1(56/gT) with the vertical.
 * Now just calculate the two angles and find their difference. If you use the value of 9.8 for g then gT turns out to be a convenient whole number. Gandalf61 (talk) 09:24, 29 July 2011 (UTC)