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16t^2+vt+s
A ball is thrown straight down from the top of a 220foot building with a velocity of 22feet per second. What is its velocity after falling 108 ft?
 2 years ago
 2 years ago
16t^2+vt+s A ball is thrown straight down from the top of a 220foot building with a velocity of 22feet per second. What is its velocity after falling 108 ft?
 2 years ago
 2 years ago

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cyterBest ResponseYou've already chosen the best response.0
\[\huge v_{f}^2 = v_{i}^2+2ad\] d=108ft a=9.8m/sec^2 vi=22ft/sec
 2 years ago

DockworkerBest ResponseYou've already chosen the best response.1
s(t)=16t^222t+220 This function measures how far above the ground the ball is, so when it falls 108 ft, it is 112 ft above the ground s(t)=16t^222t+220=112 s(t)=16t^222t+108=0 Using quadratic equation to solve for t: \[t=\frac{22 \pm \sqrt{4844(16)(108)}}{2(16)}=\frac{22 \pm 86}{16}\] We can discard the 22+86 since it will yield a negative time. So t=(2286)/32=2 v(t)=s'(t)=32t22 v(2)=32(2)22=6422= 86ft/s
 2 years ago
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