WE shot two equal sized balls out of a shooter at 80 degrees and 10 degrees, but they didn't land near the same area like they are suppossed to. Why could this happen?
2007-10-03 10:03:06 · 4 answers · asked by miller20817 3 in Science & Mathematics ➔ Physics
What do you mean by equal size? Same diameter? Same weight? If they are the same diameter, air resistance will be the same. Why do you expect a ball launched at 80º and 10º to land in the same place? Assuming both balls are launched at the same angle, if the balls have different masses but the same diameter, the air resistance is the same for both, but the deceleration due to the air resistance will be less for the more massive ball: a = F/m. Therefore the more massive (dense) ball will travel farther.
If the balls are identical, but still don't land in the same place, then you are missing something: wind, temperature differences, and small variations in initial velocity or angle. The only way conditions can be identical in a real-life situation is if the balls are both launched at the same place and the same time.
2007-10-03 10:12:45 · answer #1 · answered by gp4rts 7 · 0⤊ 0⤋
Because the size is irrelevant to where it will land (if air resistance is made negligible).
What determines displacement is the angle the object was thrown from and how fast it was sent flying.
2007-10-03 17:17:18 · answer #2 · answered by GP99 2 · 0⤊ 0⤋
Nothing really works the way they do on paper. Specially the problem of air is not taken in to account. If you actually do that .. you will find that .. the distances will not match up...
2007-10-03 17:12:26 · answer #3 · answered by Arther D 2 · 0⤊ 0⤋
Two words: Air resistance.
2007-10-03 17:06:08 · answer #4 · answered by Brent L 5 · 0⤊ 0⤋