The question here is how long does it take for a falling
person to reach the 90% of this terminal velocity. The computation is:
The terminal velocity vt fulfills v'=0. Therefore vt=g/c,
and so c=g/vt = 10/(100*1000/3600) = 36,000/100,000... /s. Incorporating the
differential equation shows that the time needed to reach velocity v is
t= ln [g / (g-c*v)] / c.
With v=.9 vt =.9 g/c,
t = ln [10] /c = 6.4 sec.
In the first quadrant the signs would be (x,y) in the second quadrant they would be (-x,y) in the third (-x,-y) and the fourth (x,-y)
Answer:
763.41 ft
Step-by-step explanation:
<h3>V=π(d/
2)2*h</h3><h3>use formula</h3>
8y+4 = 2(y-1)
8y+4 = 2y-2
8y-2y+4 = -2
6y+4 = -2
6y = -2-4
6y = -6
y = -1
It is 26 I have worked it out 278 times and it is 100% correct
