A skydiver's height above ground is

Question

A skydiver's height above ground is

h(t) = 10850 − 170t − 850e^−0.2t

where h is in feet and t is in seconds. At what instant in time is she falling at 120 feet per second? Be accurate to two decimal places and include correct units.

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Answer 1

Given:

h(t) = 10850 − 170t − 850e^−0.2t

Find:

t = ? when h'(t) = -120

Solution:

h'(t) = 170 exp( -0.2 t ) - 170

-120 = 170 exp( -0.2 t ) - 170

50 = 170 exp( -0.2 t )

exp( +0.2 t ) = 17/5

0.2 t = ln 17 - ln 5

∴ t = 5 ( ln 17 - ln 5 ) = 5 ln( 17 / 5 ) ≈ 6.119 s

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Answer 2

v(t) = h'(t) = - 170 +170e^-0.2t = 120; 170e^-0.2t = 290; t = ln(290/170)//(-0.2) = -2.67 sec

Sign "-" means that 2.67 sec ago