Answer:
7.4secs
Step-by-step 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
Answer:
The time when the swim team member is 5 feet above the water surface is 7.4 seconds.
Step-by-step 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:
[tex]\frac{-b+-\sqrt{b^{2}-4*a*c } }{2*a}[/tex]
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
[tex]\frac{-5.25+-\sqrt{5.25^{2}-4*(-0.875)*9 } }{2*(-0.875)}[/tex]
[tex]\frac{-5.25+-\sqrt{27.56+31.5 } }{-1.75}[/tex]
[tex]\frac{-5.25+-\sqrt{59.06 } }{-1.75}[/tex]
[tex]\frac{-5.25+-7.685}{-1.75}[/tex]
Then:
[tex]t1=\frac{-5.25+7.685}{-1.75}[/tex] and [tex]t2=\frac{-5.25-7.685}{-1.75}[/tex]
Solving for t1:
[tex]t1=\frac{2.435}{-1.75}[/tex]
t1= -1.39 ≅ -1.4
Solving for t2:
[tex]t2=\frac{-12.935}{-1.75}[/tex]
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.
