SOLUTION: This is a quadratic function word problem. Solve for maximum height and graph. If a baseball is projected upward from ground level with an initial velocity of 64 feet per se

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Question 23655: This is a quadratic function word problem.
Solve for maximum height and graph.
If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s(t) = -16t^2 + 54t. Graph this function for 0 ≤ t ≤ 4. What is the maximum height reached by the ball?
This problem is very confusing to me and any help will be greatly appreciated.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First: The general form of the equation for the height as a function of time of an object propelled upwards is given by: h%28t%29+=+-16t%5E2+=+vot+%2B+ho where: vo is the initial velocity at which the object is propelled and ho is the initial height from which the object is propelled.
In your problem, since the ball (object) is propelled from the ground, the initial height (ho) is zero and the initial velocity of the ball is given as 64 feet/second so vo = 64. Your equation should read:
s%28t%29+=+-16t%5E2+%2B+64t
The graph looks like this:
graph%28300%2C200%2C-5%2C5%2C-10%2C75%2C-16x%5E2%2B64x%29
Since the curve is a parabola and it opens downward, the maximum height is to be found at the vertex of the parabola.
The x-coordinate (this really corresponds to t in your equation) is given by: t+=+-b%2F2a and in your equation: s%28t%29+=+-16t%5E2+%2B+64t a = -16 and b = 64. So the t-coordinate of the vertex is:
t+=+-64%2F2%28-16%29
t+=+-64%2F-32
t+=+2 So the maximum height is attained at time t = 2 seconds.
To find the value of the height at this time, substitute t=2 into the original equation and solve for s.
s%282%29+=+-16%282%29%5E2+%2B+64%282%29
s%282%29+=+-64+%2B+128
s%282%29+=+64 feet. This is the maximum height.