Suppose h(t) = -3t^2 + 12t + 4 is an expression giving the height of a diver above the water (in meters), seconds after the diver leaves the springboard.
a. How high above the water is the springboard? Explain how you know.
b. When does the diver hit the water?
c. At what time on the diver's descent toward the water is the diver again at the same height as the springboard?
d. When does the diver reach the peak of the dive?
d.
\(h(t) = -3t^2 + 12t + 4\)
\(h'(t)=-6t+12=0\)
\(t=2\)
The diver reaches the top of the dive after 2 seconds.
c.
Up = down
\(2\times 2s=4s\)
After 4 seconds, the diver is back on the same level as the diving board.
a.
After 4 seconds, the diver is back on the same level as the diving board.
So
\(h(t) = -3t^2 + 12t + 4= -3*16 + 12*4 + 4=4\)
The board is 4m high.
b.
\(h(t) = -3t^2 + 12t + 4=0\)
\(\large x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\large x = {-12 \pm \sqrt{144-4*(-3)*4} \over -6}=4.309\)
The diver hits the water after 4.309 seconds.
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