Question 24543: suppose you throw a basebal straight up at a velocity of 64 ft per second.A funtion can be created by expressing distance above the ground, s , as a function of time, t. this function is  (<
16 represents 1/2 g,gravitational pull in ft per second^2.
v0is the inital velocity in ft per second
so is intial distance above ground in ft. if you are stading on the ground , then s0= zero.
what is the function descibing this problem?
how high above the ground will the ball be after one sec?
how long will it take to hit the ground?
what is the maximum height and time attained by the ball?
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Hmmm...I think I have answered this one previously. Anyway, here it is again.
The equation is correct:
 This is the function describing the problem. In words, the height (s) is a quadratic function of time (t) with initial velocity of 64 ft/sec and an initial height of zero.
After 1 second (t=1), the height of the baseball will be:
 The height will be 48 feet after 1 second.
Find when the baseball will hit the ground (s=0) by setting the function s(t) = 0 and solving for t.
 Factor out a t.
 Divide both sides by t.
 Subtract 64 from both sides.
 Divide both sides by -16
 The baseball will return to ground in 4 seconds.
The maximum height and time is found by finding the location of the vertex of this parabola.
Since the parabola opens downward, the vertex will be at the maximum point of the vertex and will represent the maximum height attained by the baseball.
The x-coordinate (or, in this problem, the t-coordinate) is given by:
The a and b come from the standard form of the quadratic equation: 
In this case, a = -16, b = 64, and c = 0
 The maximum height of the baseball will be attained in 2 seconds.
To find this maximum eight, substitute t=2 into the original function and solve for s.
 The maximum height attained by the baseball is 64 feet.
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