At the pool, Janae jumps off a diving board that is 15 feet high. Janaes height above the water is modeled by the function f(x)=-5x^2+10x+15. 1.Janae made a perfect dive and entered the water hands first. After she surfaced and got out of the water, Janae wanted to know when she hit the water. HOW CAN JANAE FIND OUT WHEN SHE FIRST HUT WATER(how long was she in the air)?2. How long was Janae in the air before she finished her dive and splashed into the water? There are several ways to determine this answer but you must explain all steps on how u found the solution.

Question

27-10-2022
Mathematics

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At the pool, Janae jumps off a diving board that is 15 feet high. Janaes height above the water is modeled by the function f(x)=-5x^2+10x+15. 1.Janae made a perfect dive and entered the water hands first. After she surfaced and got out of the water, Janae wanted to know when she hit the water. HOW CAN JANAE FIND OUT WHEN SHE FIRST HUT WATER(how long was she in the air)?2. How long was Janae in the air before she finished her dive and splashed into the water? There are several ways to determine this answer but you must explain all steps on how u found the solution.

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LibnyY788421 LibnyY788421 · Answer 1 · 2022-10-27T13:37:00-04:00

Question 1 : Janae can find out when she first hit the water by finding out the distance of travel, as well as her speed of travel

Question 2 :

Given that her height above the water can be modeled using the function below:

[tex]f(x)=-5x^2\text{ + 10x + 15}[/tex]

We can obtain her vertical distance of travel from the diving board to the point where she starts to descend

At the turning point,

[tex]\frac{df(x)}{dx}\text{ = 0}[/tex][tex]\begin{gathered} \frac{df(x)}{dx}\text{ = 10x + 10 = 0} \\ x\text{ = -1} \\ \text{substituting back into f(x)} \\ f(-1)=5(-1)^2\text{ + 10(-1) + 15} \\ =\text{ 5 - 10 + 15} \\ =\text{ 10} \end{gathered}[/tex]

Hence, Jane travelled a distance of 10 units upwards and (10 + 15)unit downwards

At the turning point, her velocity is zero, using the relation below we can find her initial velocity and then the time it took

[tex]\begin{gathered} v^2=u^2\text{ - 2gS} \\ 0=u^2\text{ - 2 }\times\text{ 10 }\times10 \\ u\text{ = 14.14 m/s} \\ v\text{ = u - gt } \\ 0\text{ = 14.14 - 10 }\times\text{ t} \\ t\text{ = 1.414s } \end{gathered}[/tex]

From the turning point, her velocity changes from 0