A water balloon launched from the roof of a building at time t = 0 seconds has vertical velocity v(t) = -32t + 40 feet/sec with positive v corresponding to upward motion.
s(t) = -16t^2+40t+60
A 6-foot person is standing on the ground. How fast is the water balloon falling when it strikes the person on the top of the head? Give an exact value.
speed = ? feet / sec
Differential equation
2008-12-04 12:12 pm
回答 (1)
2008-12-04 4:48 pm
✔ 最佳答案
When t = 0, s = 60, that is the balloon is 60 ft above ground. So head of the person is 60 - 6 = 54 ft above ground. For s = 54, we get
54 = - 16t^2 + 40t + 60
16t^2 - 40t - 6 = 0
8t^2 - 20t - 3 = 0
t = [20 +/- sqrt(400 + 96)]/16= (5 +/- sqrt31)/4.
since t is positive so t = (5 + sqrt31)/4.
Sub into the equation of v, we get
v = -8(5 + sqrt31) + 40
= -8sqrt31, negative v means downward motion.
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