At t = 0, a stuntwoman jumps off the top of a tall building. She is wearing a sp

Question

At t = 0, a stuntwoman jumps off the top of a tall building. She is wearing a special suit, which increases the air resistance she experiences so that, after falling with constant acceleration for a time of t1 = 4.00 s, she then falls with a constant velocity of 20 m/s directed down. At a time of t2 = 22.0 s after leaving the top of the building, the stuntwoman reaches a safety net, which provides a constant upward acceleration that brings the stuntwoman to rest at a time of t3 = 24.0 s after leaving the top of the building. The graph below shows the vertical component of the stuntwoman's velocity throughout the entire process. See if you can use the graph to help you solve the problem.



Assuming the stuntwoman comes to rest at ground level (this was a very carefully planned stunt!), how tall is the building? For reference, the John Hancock Tower in Boston, the tallest building in New England, is 241 m tall.

Explanation / Answer

assume the acceleration from t=0 to t1is a1

v1 = 0 + a1t1

a1 = v1/t1 = 20/4 = 5.0 m/s2

distance traveled in this time period

s1 = 1/2 a1t12 = (1/2)(5.0)(4)2 = 40 m

distance traveled from t1 to t2

s2 = v1(t2-t1) = 20 (22-4) = 360 m

assume the acceleration from t2 to t3 is a2, then

0 = v1+ a2(t3-t2)

a2 = -20 /(24-22) = -10 m/s2

distance traveled in this time period

s3 = v1(24-22) + (1/2)(-10)(24-22)2 = 20*2 -20 = 20 m

so the build height is

s1+s2+s3 = 40+360+20 = 420 m

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