Diving into Fun Calculations with Diver's Heights

a. How high above the water is the springboard?

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b. When does the diver hit the water?

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c. At what time on the diver's descent toward the water is the diver again at the same height as the springboard?

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d. When does the diver reach the peak of the dive?

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Final Answer:

The springboard is at 0 meters. The diver hits the water at t=3.79 seconds and passes the springboard's height again at the same time. The diver reaches the peak of the dive at t=2.424 seconds.

To analyze the diver's motion, we'll use the given formula h(t) = -4t² + 11t³. First, the height of the springboard is the value of h(0), which equals -4*0² + 11*0³ = 0 meters. When the diver hits the water, the height h(t) is equal to zero. We can solve the equation -4t² + 11t³ = 0 for t, acquiring roots t=0 and t=3.79. As t=0 represents the beginning of the dive, thus the diver hits the water at t=3.79 seconds.

To find when the diver comes back to the springboard's height, as the springboard's height is 0 meters, we have the same equation -4t² + 11t³ = 0, hence the diver is at the same height as the springboard again at t=0 and at t=3.79s on the way down. The peak of the dive is attained when the rate of change of height equals zero. This implies the derivative of the height function (h'(t)= -8t + 33t²) needs to be equal to zero which gives us two times: t=0 and t=2.424. The second is the correct time as it falls in between of the fall, therefore the diver reaches the peak of the dive at t = 2.424 seconds.