Answer:
it so easy that correct
Step-by-step explanation:
$500.00×$8.5
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.
Speed=distance/time
suppose the distance of the first day is d, and the time is t
distance of the second day: d+0.17d=1.17d
time of the second day: t+0.2t=1.2t
speed of the second day: 1.17d/1.2t=0.975(d/t)=(1-0.025)(d/t)
so the speed of the second day is 2.5% slower than the first day.
