Diving into Fun Calculations with Diver's Heights
a. How high above the water is the springboard?
Show your work.
b. When does the diver hit the water?
Show your work.
c. At what time on the diver's descent toward the water is the diver again at the same height as the springboard?
Show your work.
d. When does the diver reach the peak of the dive?
Show your work.
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.