Question

Hayden is standing on a ledge that is 32 feet above a mountain stream. At time t = 0, he jumps from the ledge into the water below. Hayden's position at any time t ≥ 0 can be modeled by the function s(t) = -16t² + 16t + 32, where s is measured in feet, and t is measured in seconds. a. When does Hayden hit the water? b. What is Hayden's velocity at impact?

Answer 1

a. Hayden hits the water at 2 seconds. b. Hayden's velocity at impact is -48 feet per second.

Step-by-step explanation:

a. To find when Hayden hits the water, we need to find the time when his position is at the same level as the water, which is at a height of 0 feet. So we set s(t) = 0 and solve for t.
-16t² + 16t + 32 = 0
This is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. The solutions are t = 2 seconds and t = -1 second. Since time cannot be negative, the answer is t = 2 seconds.
b. To find Hayden's velocity at impact, we need to find the derivative of the position function s(t). The derivative of -16t² + 16t + 32 is v(t) = -32t + 16. So, plugging in t = 2 into the velocity function gives us v(2) = -32(2) + 16 = -48 feet per second. Therefore, Hayden's velocity at impact is -48 feet per second.