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Question

A cliff on a Caribbean island jumps from a height of 105 feet , with an initial upward velocity of 5 feet per second. An equation that models the height , h(t) , above the water , in feet , of the diver in time elapsed,t, in seconds ,is h(t)=-16t²+5t +105 . How many seconds, to the nearest hundredth, does it take to diver to fall 45 feet below the starting point ?

Answer 1

h = 105 ft - 45 ft = 60 ft

h(t) = -16t² + 5t + 105

60 ft = -16t² + 5t + 105

0 = -16t² + 5t + 45

Use the quadratic formula to solve for t:

t = [-5 ± √(5² - 4(-16)(45))] / 2(-16) = [-5 ± √2905] / (-32) = -1.53, 1.84

Reject the negative answer:

t = 1.84 s

Answer 2

45 ft below of startin point means that h(t) = 105 - 45 = 60 ft. So we get equation -16t² + 5t + 105 = 60 or

16t² - 5t - 45 = 0. From here t = 1.841 sec