Problem Solving-Quadratic Formula. Don't quite understand how to solve

Question

In a shot put even, Jenna tosses her last shot from a position of about 6 feet above the ground with an initial vertical and horizontal velocity of 20 ft./s the height of the shot is modeled by the function H(t)=-16t^2+20t+6, where T is the time in seconds after the toss. The horizontal distance traveled after T seconds is mod by d(t)=20t

 

1. jenna wants to know the exact distance the shot travels at a velocity of 20 feet per second.

A. Use Quadratic Formula T=(-b±√B^2-4ac)/2a to solve height function for t.

B. Use the value for t and the distance function to find the distance her shot travels.

 

2. (Need most help with) Jenna is working to improve her performance. she makes a table to show how the horizontal distance varies with velocity. Complete the table.

 

Velocity (ft/s) Formula

22.          -22±√(22)^2-4(-16)(6)

25.           -blank-

28.            -blank-

Time                Distance

-blank-             -blank-

-blank-             -blank-

-blank-             -blank-

Answer 1

You have to use the distance function to find how

far the shot put traveled.   In this function, you 

have the variable t which represents the time it 

was in the air.  You find t in the Height equation.

 

The height of the shot put when it has traveled 

its maximum distance will be 0 because

the shot put will be on the ground.

 Therefore:

 

          H(t) = - 16t² + 20t + 6

 

              0 = -16t² + 20t + 6

 

Use the quadratic formula to solve for t:

 

     t = -20 ± √[ 20² -4(-16)(6)] / 2(-16)

 

     t = (-20 ± 28) / -32

 

t = -.25 sec ;   1.5 sec

 

Disregard the -.25 sec

 

d(1.5 sec) = 20 (1.5) = 30 feet traveled horizontally

Answer 2

Hi Athena.

 

In your quadratic function, your velocity will be denoted by the "b" term. You'll notice in your first equation, where your velocity was 20, your function was given by:

 

H(t) = -16t^2 + 20t + 6

 

With t equaling:

 

t = [ -20 + √(20^2 - 4(-16)(6)) ] / (-2*16)

 

And in your second equation, where the velocity is given as 22, your equation becomes:

 

H(t) = -16t^2 + 22t + 6

 

With t equaling:

 

t = [ -22 + √(22^2 - 4(-16)(6)) ] / (-2*16)

 

Following this pattern, you can determine what the t value is for a velocity of 25 and 28.

 

Once you know the t for each, distance is given by:

 

v*t

 

So once you find t_20 for a velocity of 20, your distance will be:

 

20*t_20

 

Likewise, once you find t_22 for a velocity of 22, your distance will be:

 

22*t_22

 

And so on and so forth.

 

Let me know if anything isn't making any sense. Hope this helps.