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a diver jumps from a platform diving board that is 32 ft above the water......?
At time t = 0, a diver jumps from a platform diving board that is 32 ft above the water. The position of the diver is given by the function
s(t)= -16t^2+16t+32
where s is measured in feet and t is in seconds.
(i) When does the diver hit the water?
(ii) What is the diver’s velocity on impact?
3 Answers
- CwCcLv 71 decade agoFavorite Answer
s(t)= -16t^2+16t+32=0
t=-1 or 2; has to be 2 s
s'(t) = v(t) = -32t+16
v(2) = - 48 ft/s [negative denotes down]
- JohnLv 71 decade ago
Factor s(t) first.
s(t) = -16(t^2 - t - 2) = -16(t - 2)(t + 1).
(i) He hits the water 2 seconds after he leaves the board.
(ii) Velocity on impact is s'(2) = -32t + 16 = -32(2) + 16 =
-48 feet/second.
-John
