Question

A competitive diver dives from a 33-foot high diving board. The height of the diver in feet after 't' seconds is given by u(t) = −16t^2 + 4t + 33. At the moment the diver begins her dive, another diver begins climbing the diving board ladder at a rate of 2 feet per second. At what height above the pool deck do the two divers pass each other? Please answer it quickly, it's for my homework.

Answer 1

Answer:

$t = 0.375s$

Step-by-step explanation:

Given

$h(t) = -16t^2 + 4t + 33$ --- driver 1

$Rate = 2ft/s$ -- driver 2

$height = 33ft$

Required

The time they passed each other

First, we determine the function of driver 2.

We have that:

$Rate = 2ft/s$ and $height = 33ft$

So, the function is:

$h_2(t) = Height - Rate * t$

$h_2(t) = 33 - 2t$

The time they drive pass each other is calculated as:

$h(t) = h_2(t)$

$-16t^2 + 4t + 33= 33 - 2t$

Collect like terms

$-16t^2 + 4t + 2t= 33 - 33$

$-16t^2 + 6t= 0$

Divide through by 2t

$-8t + 3= 0$

Solve for -8t

$-8t = -3$

Solve for t

$t = \frac{-3}{-8}$

$t = 0.375s$