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?
At t= 0 the sprin board is 4m above the water. Check the value of the equation with t=0.
The diver hits the water at t=4.30 seconds, approximately. This is the zero of the formula. Check with the quadratic formula.
The diver is at the springboard height again at 4 seconds. Set the equation = to 4 and check that.
The peak of the dive is at 2 seconds. Here you use the first derivative and find the zero for that equation.
That zero is the inflection point where the curve reaches a maximum value. It goes down from there.
First derivative is equal to -6t + 12.
+14986 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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