When we solve free-fall problems near Earth, it\'s important to remember that ai

Question

When we solve free-fall problems near Earth, it's important to remember that air resistance may play a significant role. If its effects are significant, we may get answers that are wrong by orders of magnitude if we ignore it. How can we tell when it is valid to ignore the effects of air resistance? One way is to realize that air resistance increases with increasing speed. Thus, as an object falls and its speed increases, its downward acceleration decreases. Under these circumstances, the object's speed will approach a limit, a value called its terminal speed. This terminal speed depends upon such things as the mass and cross-sectional area of the body. Upon reaching its terminal speed, its acceleration is zero. For a "typical" skydiver falling through the air, a typical terminal speed is about 49.7 m/s (roughly 111 mph). At half its terminal speed, the skydiver's acceleration will be about 3/4 g. Let us take half the terminal speed as a reasonable "upper bound" beyond which we should not use our constant acceleration free-fall relationships. o, the skydiver als before we can no longer neglect air resistance. (a) Assuming the skydiver started from rest, estimate how far, and for how long, the skydiver falls before we can no longer neglect air resistance. 14.97 (b) Repeat the analysis for a Ping-Pong ba, which has a terminal speed of about 5.4 m/s y = 148.7 (c) What can you conclude by comparing your answers for Parts (a) and (b)? air resistance is more for light e is more for lighter is ping - bong bal

Explanation / Answer

5)
a = 9.8 m/s^2
t = t
d = H
Vi = 0

use:
d = Vi*t + 0.5*a*t^2
H = 0 + 0.5*9.8*t^2
t = sqrt (2*H/9.8) .....eqn 1

Now calculate distance travelled in t-1 s
use:
d = Vi*t + 0.5*a*(t-1)^2
= 0 + 0.5*9.8*(t-1)^2
= 4.9*(t-1)^2

distance travelled in last second = H - d
= H - 4.9*(t-1)^2
so,
34.7 = H - 4.9*(t-1)^2
(t-1)^2 = (H-34.7)/4.9
t^2 - 2*t + 1 = (H-34.7)/4.9
2*H/9.8 - 2* sqrt (2*H/9.8) + 1 = (H-34.7)/4.9

Let sqrt(H) = h
then H = h^2

2*h^2/9.8 - 2* h*sqrt (2/9.8) + 1 = (h^2-34.7)/4.9
0.204h^2 - 0.904*h + 1 = 0.204*h^2 - 7.082
0.904*h = 8.082
h = 8.94 m

so,
sqrt(H) = h
sqrt(H) = 8.94
H = 80 m
Answer: 80 m

6)
Vf^2 = Vi^2 + 2*a*d
0 = Vi^2 + 2*(-7.5)*4.1
Vi = 7.84 m/s
Answer: 7.84 m/s

Answered 2 of your question.Thanks

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