SOLUTION: A baseball is hit 2 feet above the ground with initial upward velocity of 48 feet per second. s(t)=-16t^2+42t+2 where the heigth is, s(t), is in feet and the time t is in seconds.

Algebra ->  Length-and-distance -> SOLUTION: A baseball is hit 2 feet above the ground with initial upward velocity of 48 feet per second. s(t)=-16t^2+42t+2 where the heigth is, s(t), is in feet and the time t is in seconds.       Log On


   

Question 1135964: A baseball is hit 2 feet above the ground with initial upward velocity of 48 feet per second. s(t)=-16t^2+42t+2 where the heigth is, s(t), is in feet and the time t is in seconds. A) at what time does the ball reach the maximum height? B)What is the maximum height of the ball? C)when does the ball hit the ground for the first time? Use A,B,C to Graph the parabola


Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.

            When you see the problem like this one, REMEMBER:

            The maximum height is the maximum value of the given quadratic function.

            For the quadratic function  y = ax^2 + bx + c   (with the negative leading coefficient)  the plot is a downward parabola,

                    and it has the maximum at   x = -b%2F%282a%29. 


(A)  In your case the given quadratic function gets the maximum at  t = -42%2F%282%2A%28-16%29%29 = 42%2F32 = 21%2F16 seconds = 15%2F16 seconds.



(B)  The maximum height is the value of this quadratic function at t= 21%2F16 seconds

    s%2821%2F16%29 =  -16%2A%2821%2F16%29%5E2+%2B+42%2A%2821%2F16%29+%2B+2 = 29.56 ft.




(C)  The ball will hit the ground for the first time when s(t) = 0.


     To find this time, you need to solve this quadratic equation


          -16t%5E2+%2B+42t+%2B+2 = 0,    or, equivalently


          8t%5E2+-+21t+-+1 = 0


          t%5B1%2C2%5D = %2821+%2B-+sqrt%2821%5E2+-+4%2A8%2A%28-1%29%29%29%2F%282%2A8%29 = %2821+%2B-+sqrt%28473%29%29%2F16 = %2821+%2B-+21.748%29%2F16.


          Use the positive root only  t = %2821+%2B+21.748%29%2F16 = 2.672 seconds.


The plot is shown below :



    graph%28+330%2C+330%2C+-1%2C+4%2C+-5%2C+40%2C%0D%0A++++++++++-16x%5E2%2B42x%2B2%0D%0A%29


         Plot y = -16x%5E2%2B42x%2B2

All questions are answered and explained - the problem is solved.

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To see other similar solved problems,  look into the lessons
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

Consider these lessons as your textbook, handbook, tutorials and  (free of charge)  home teacher.
Read them attentively and learn how to solve this type of problems once and for all.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.