Question

-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?

Answer 1

\[\huge v_{f}^2 = v_{i}^2+2ad\]
d=108ft
a=-9.8m/sec^2
vi=-22ft/sec

Answer 2

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