Question 51271: I am confused on how to solve this problem.
Suppose a baseball is shot up from the ground straight up with an initial velocity of 32 feet per second. A function can be created by expressing distance above the ground, s, as a function of time, t. This function is s = -16t2 + v0t + s0
• 16 represents 1/2g, the gravitational pull due to gravity (measured in feet per second2).
• v0 is the initial velocity (how hard do you throw the object, measured in feet per second).
• s0 is the initial distance above ground (in feet). If you are standing on the ground, then s0 = 0.
What is the maximum height of the ball? What time will the maximum height be attained?
I wasn't sure how to do this. I want to say 32 feet. But I'm not sure.
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Let's answer all three questions for the problem here.
Since the initial upward velocity is given as 32 ft/sec, you can write the function of height,s, as a function of time, t. If you were to graph this function, you would get a parabola that opens downward. We know that because the coefficient of the t^2 term is negative.
 For question #2, the time at which maximum height is attained by the baseball can be found by locating the vertex of the parabola then finding the value of t that corresponds to this location.
The t-coordinate of the vertex is given by:  .
Where do the a and b come from?
They come from comparing your equation with the general standard form of the quadratic equation:  but in your equation of course, the independent variable is t rather than x and a = -16 and b = 32.
So, let's find the the value of t at which the vertex is located.
In your equation, a = -16 and b = 32. Substituting these values into the appropriate places, we get:
 Simplifying this:
 second.
The baseball attains its maximum height in 1 second.
For question #1, the maximum height attained by the baseball can be found by substituting t = 1 in the original function for the height:  and solveing for the height, s.
 Simplifying, we get:
 feet.
The maximum height attained by the baseball is 16 feet.
Question #3 (Posted separately) At what time does the baseball return to the ground? Well, that's really asking at what time is the height of the base = 0 feet. There will, of course, be two anwers to this question. The height will be zero at the start of the action and then it will be zero when the baseball returns to earth, so only the second solution is of interest to us. To find this time, we'll set the function for the height equal to zero and solve for the time, t.
 Factor a t.
 Apply the zero product principle.
 and/or 
Well, t = 0 is the first solution whch we can ignore.
 Subtract 32 from both sides of the equation.
 Divide both sides by -16.
 seconds.
The baseball returns to the ground at 2 seconds.
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