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

Mathematics
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\[\huge v_{f}^2 = v_{i}^2+2ad\] d=108ft a=-9.8m/sec^2 vi=-22ft/sec
s(t)=-16t^2-22t+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^2-22t+220=112 s(t)=-16t^2-22t+108=0 Using quadratic equation to solve for t: \[t=\frac{22 \pm \sqrt{484-4(-16)(108)}}{2(-16)}=\frac{22 \pm 86}{-16}\] We can discard the 22+86 since it will yield a negative time. So t=(22-86)/-32=2 v(t)=s'(t)=-32t-22 v(2)=-32(2)-22=-64-22= -86ft/s

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