Question

A swim team member performs a dive from a 14-foot high springboard. Her path is represented by the equation LaTeX: h\left(t\right)=-0.875t^2+5.25t+14h ( t ) = − 0.875 t 2 + 5.25 t + 14, where t is time in seconds and h is the height above the surface of the water. She needs to finish her flip no later than 5 feet above the surface of the water in order to prepare for the splash. What time does this happen? Round to the nearest tenth of a second.

Answer 1

The time when the swim team member is 5 feet above the water surface is 7.4 seconds.

Explanation:

You want to know when a member of the swim team finished his jump no later than 5 feet above the surface of the water to prepare for the splash, that is, when t is 5 feet above the surface of the water. Substituting h(t) for this value 5 in the equation you obtain:

5= -0.875*t²+5.25*t+14

A quadratic equation has the general form:

a*x² + b*x +c= 0

where a, b and c are known values ​​and a cannot be 0.

Taking the equation 5= -0.875*t²+5.25*t+14 to that form, you get:

-0.875*t²+5.25*t+14-5=0

-0.875*t²+5.25*t+9=0

The roots of a quadratic equation are the values ​​of the unknown that satisfy the equation. And solving a quadratic equation is finding the roots of the equation. For this you use the formula:

`\frac{-b+-\sqrt{b^(2)-4*a*c } }{2*a}`

In this case, solving the equation is calculating the values ​​of t, that is, you find the time when the swim team member is 5 feet above the water surface.

Being a= -0.875, b=5.25 and c= 9, then

`\frac{-5.25+-\sqrt{5.25^(2)-4*(-0.875)*9 } }{2*(-0.875)}`

`(-5.25+-√(27.56+31.5 ) )/(-1.75)`

`(-5.25+-√(59.06 ) )/(-1.75)`

`(-5.25+-7.685)/(-1.75)`

Then:

`t1=(-5.25+7.685)/(-1.75)` and
`t2=(-5.25-7.685)/(-1.75)`

Solving for t1:

`t1=(2.435)/(-1.75)`

t1= -1.39 ≅ -1.4

Solving for t2:

`t2=(-12.935)/(-1.75)`

t2= 7.39 ≅ 7.4

Since time cannot be negative, the time when the swim team member is 5 feet above the water surface is 7.4 seconds.

Answer 2

7.4secs

Explanation:

If the path of a swim team member is modeled by the equation

h(t) = -0.875t²+5.25t+14

where t is time in seconds and h is the height above the surface of the water

If she need to finish in no later than 5feet above the surface, the expression will become:

-0.875t²+5.25t+14≤5

-0.875t²+5.25t+14-5≤0

-0.875t²+5.25t+9≤0

Rewriting using fraction

-7/8t²+21/9t+9≤0

To get the time that this happens we will need to factorize the resulting expression above;

On factorising

Find the roots of the equation using the formulas

t = 3+3√105/7

t = 3+ 3(10.25)/7

t = 3 + (30.75)/7

t = 3 + 4.393

t = 7.393

t = 7.4secs

Hence this happens at t = 7.4secs