The terminal velocity of a person falling through air is about 100km/hr. the acceleration due to gravity is 10ms−2. use this information to find out what c is in the equation v′=c(gc−v), where g is the acceleration due to gravity and v is the velocity.

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.

the distance from any point on the parabola to the focus is the exact same as the distance from that point to the directrix.

if you could provide answer choices that might allow me to answer the question easier if my above answer isn't an option! (:

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.