physics 拋體運動一問

2011-08-12 5:57 am
球以投射角30度向上飛出 , u是5 ms-1. 球下降至原本位置的高度時 , v=?

回答 (2)

2011-08-14 3:48 am
✔ 最佳答案
如果相同高度既話u=v
考慮水平分量速率,假設冇阻力,是勻速運動
考慮垂直分量速率,假設冇阻力,在同一點高度的速率是相等
跟據v^2=u^2 + 2as
垂直分量的位移為零,所以v=u

總合水平及垂直分量v都是等於u

所以v=5 ms-1
2011-08-12 6:49 am
Assume air resistance is negligible. The speed v is 5 m/s. This is a result from the Law of Conservation of Energy.

Since the levels at the starting point and end point are the same, the potential energies at these two points are equal. Because the sum of kinetic and potential energies at any point on the projectile path is constant (Law of Conservation of Mechanical Energy), equal potnetial energy at the starting and end points indicates equal kinetic energy. Hence, the speeds of the ball are equal.


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