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.

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MrRoyal MrRoyal · Answer 1 · 2021-05-29T03:49:50+02:00

Answer:

[tex]t = 0.375s[/tex]

Step-by-step explanation:

Given

[tex]h(t) = -16t^2 + 4t + 33[/tex] --- driver 1

[tex]Rate = 2ft/s[/tex] -- driver 2

[tex]height = 33ft[/tex]

Required

The time they passed each other

First, we determine the function of driver 2.

We have that:

[tex]Rate = 2ft/s[/tex] and [tex]height = 33ft[/tex]

So, the function is:

[tex]h_2(t) = Height - Rate * t[/tex]

[tex]h_2(t) = 33 - 2t[/tex]

The time they drive pass each other is calculated as:

[tex]h(t) = h_2(t)[/tex]

[tex]-16t^2 + 4t + 33= 33 - 2t[/tex]

Collect like terms

[tex]-16t^2 + 4t + 2t= 33 - 33[/tex]

[tex]-16t^2 + 6t= 0[/tex]

Divide through by 2t

[tex]-8t + 3= 0[/tex]

Solve for -8t

[tex]-8t = -3[/tex]

Solve for t

[tex]t = \frac{-3}{-8}[/tex]

[tex]t = 0.375s[/tex]