An archer is shooting an arrow at a target directly to the west. The arrow leaves the bow at 31.0 m/s. There is a wind blowing 25.0 degrees east of south at 26.0 m/s. The archer forgets to compensate for the wind. What was the velocity (including and and direction) of the arrow?
2006-10-27 16:34:04 · 7 answers · asked by h_oboe 2 in Science & Mathematics ➔ Physics
Can you tell me how you solved it??
2006-10-27 16:40:48 · update #1
You need to decompose the wind velocity into an east component and a south component. The east component is in opposition to the westward velocity of the arrow and must be subtracted from the arrow's westward velocity. That resultant is then vector summed with the southward velocity. The arctangent of the westward velocity divided by the southward velocity is the angle south of west.
2006-10-27 16:52:41 · answer #1 · answered by arbiter007 6 · 0⤊ 0⤋
Well... this is a pretty simple problem. You just have to add the vectors together. (As long as you are assuming that you are neglecting gravity and air friction).
Draw a picture. Make an arrow pointing directly west. Now... this part depends on how your problem is presented... is it blowing to east of south or from east of south? Once you figure that out, you draw another arrow representing the wind, you draw the tail from the head of the one you made for the arrow. Now you complete the triangle by drawing a third arrow from the head of the wind back to the tail of the first arrow. Look at that, you now know two magnitudes and you know the angle between them. Use the law of cosines to figure out the rest.
2006-10-27 17:10:00 · answer #2 · answered by Just another 2D character online 3 · 0⤊ 0⤋
This is a problem of vectors. An arrow move to the west and the wind move to the east (assuming this two vectors work exatctly at the opposite direction).
The resultant of action of the wind should decrease the speed of the arrow.
In order to calculate how much decrease on the speed of the arrow, you need to calculate the friction force resulted by the wind. The total speed of the wind doesnot directly reduce the speed of the arrow, its depend on the frictional coefisience applied.
2006-10-27 18:29:18 · answer #3 · answered by techeroflogic 2 · 0⤊ 0⤋
At what distance? I assume the teacher wants a simple vector analysis. Use trig to calculate the x and y coordinates of the initial arrow velocity and the wind. The arrow is going in the negative x direction. The wind is going in the positive x and negative y direction. Add the x's and y's and the hypotenuse of the two will give you the final vector.
2006-10-27 16:42:10 · answer #4 · answered by Richard B 4 · 0⤊ 0⤋
sounds like u need to determine the actual impact of colliding vectors and take the resulting vector I don't believe they really have given the entire needed data for a real answer but give the academic one anyway, I suppose? just because the wind is calibrated as such doesn't explain how mauch actual affect it has on the arrow does it ?
2006-10-27 16:44:13 · answer #5 · answered by dogpatch USA 7 · 0⤊ 0⤋
Marksman 2 by way of fact he can shoot greater formerly having to reload! haha. quite nevertheless - look at it this way. how many seconds does it take each physique to fire off ONE shot? Marksman a million: 5 photographs in 5 seconds = 5/5 = a million 2d in step with shot. Marksman 2: 10 photographs in 10 seconds = 10/10 = a million 2d in step with shot. Wow! their velocity in step with shot is equivalent - so clearly the answer to 'which guy can fire 12 photographs in a shorter time' may be... NEITHER!
2016-10-16 11:52:18 · answer #6 · answered by ? 4 · 0⤊ 0⤋
29m/s south east
2006-10-27 16:38:04 · answer #7 · answered by franklino 4 · 0⤊ 2⤋