SOLUTION: A cliff driver jumps from the top of a cliff that is 150 feet tall. the distance from the top of the cliff to the water can be modeled by the function y = =- 16x² +150 where x is

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Question 1198226: A cliff driver jumps from the top of a cliff that is 150 feet tall. the distance from the top of the cliff to the water can be modeled by the function y = =- 16x² +150 where x is the time in seconds and y is the height from the ground as the diver jumps. when will the diver be at a height of 15 feet? how long will it take for the diver to splash into the water?
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Part 1) When will the diver be at a height of 15 feet?

x = elapsed time in seconds
y = height in feet

Plug in y = 15 and solve for x.
y = -16x^2 + 150
15 = -16x^2 + 150
15-150 = -16x^2
-135 = -16x^2
x^2 = -135/(-16)
x^2 = 8.4375
x = sqrt(8.4375)
x = 2.90473750965557
x = 2.9047
It takes about 2.9047 seconds to reach a height of 15 feet. In such context we ignore air resistance because it greatly complicates the problem. Realistically, the time value will be slightly larger than 2.9047 because the air slows the person down.

Answer: Approximately 2.9047 seconds

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Part 2) How long will it take for the diver to splash into the water?

The water level is at y = 0 feet in height.
y = -16x^2 + 150
0 = -16x^2 + 150
16x^2 = 150
x^2 = 150/16
x^2 = 9.375
x = sqrt(9.375)
x = 3.06186217847898
x = 3.0619
Similar to the previous part, air resistance is ignored (and realistically air resistance will slow the person down to make the time duration slightly longer than 3.0619 seconds).

Answer: Approximately 3.0619 seconds