A parachute at height h above the ground is dropped vertically downward with
initial speed V = O. Assume the total weight of the parachute plus the payload is
W and the drag can be represented by kVV where k is constant.
Determine the speed V(h) of the parachute when it hits the ground .
註:第三行 KVV是K乘以(V的平方),因我無法打出平方符號,故如此表示,謝謝!
普通物理--研究所考古題
2013-02-18 5:44 am
回答 (3)
2013-02-19 5:16 am
✔ 最佳答案
一、降落傘開始下落時，阻力會隨著 v² 快速增加，此過程W – kv² = ma -----（a為降落傘及人的向下加速度，假設向下為正）
因 kv² 不斷增大，所以 a 不斷減少，是變加速運動。
二、降落傘下落一段時間後，當「阻力＝重力」時，會變成「等速度」下落，
此時的速度稱為「終端速度」。列式如下
W – kv² = 0
所以，降落傘著地前 W = kv²
即，著地速度 v = √W/k（註：W/k 均在根號內）
2013-02-19 15:29:01 補充：
補充：為何不去計算第一項變加速度過程中的速度？
因為，開傘前，自由落體的速度可達終端速度的五倍，這樣才能確保降落傘打開。
即使，達到了終端速度，著地時一不小心就會受重傷。
所以，題目要問的是著地速度，應該是「終端速度」。
參考: 自已
2013-02-20 6:11 pm
非常謝謝Edson,DaSaGwa兩位!
2013-02-19 6:44 am
Your consideration is valid. If so, you might need to use
-kV^2 + W = m(dV/dt) to solve for V, then use a = dV/dt to get acceleration.
However, this can involve with a very complicated ODE solving.
2013-02-18 22:48:26 補充：
Edson's solution is good, however, it is not complete, he needs find out at what height, the velocity reaches constant. Hence, he has only the velocity when it reach constant, but doesn't have the velocity BEFORE it becomes constant.
2013-02-18 22:49:20 補充：
I removed my answer, because I didn't consider the acceleration is function of velocity. Sorry for my carelessness!
2013-02-18 23:07:24 補充：
Of course, to find the velocity before it reaches constant, you will need to use the differential equation.
I wonder whether the people who made the question has these concerns ?
-kV^2 + W = m(dV/dt) to solve for V, then use a = dV/dt to get acceleration.
However, this can involve with a very complicated ODE solving.
2013-02-18 22:48:26 補充：
Edson's solution is good, however, it is not complete, he needs find out at what height, the velocity reaches constant. Hence, he has only the velocity when it reach constant, but doesn't have the velocity BEFORE it becomes constant.
2013-02-18 22:49:20 補充：
I removed my answer, because I didn't consider the acceleration is function of velocity. Sorry for my carelessness!
2013-02-18 23:07:24 補充：
Of course, to find the velocity before it reaches constant, you will need to use the differential equation.
I wonder whether the people who made the question has these concerns ?
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