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taurus [48]
3 years ago
14

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.
Mathematics
1 answer:
ludmilkaskok [199]3 years ago
7 0

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

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