SOLUTION: The height in feet for a ball thrown upward at 48 feet per second is given by s(t)=-16t^2 +48 where t is the time in seconds after the ball is tossed. What is the maximum height

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Question 169248: The height in feet for a ball thrown upward at 48 feet per
second is given by s(t)=-16t^2 +48 where t is the time
in seconds after the ball is tossed. What is the maximum
height that the ball will reach?
Thank you.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You do have a small but significant error in your equation. The general form for the function of the height (as a function of time) of an object propelled upward at an initial velocity of v%5B0%5D from an initial height of h%5B0%5D is given by:
h%28t%29+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D so your equation should read:
h%28t%29+=+-16t%5E2%2B48t and, for your problem, v%5B0%5D+=+48 and h%5B0%5D+=+0.
You want to find the maximum height of the ball which means you need to know the location of the vertex (a maximimum in this case) of the parabola represented by the equation.
The t-coordinate of this parabola is given by:
t+=+-b%2F2a This will give you the time at which the ball reaches the maximum height. Once you have this, you can then find the maximum height.
t+=+-%2848%29%2F2%28-16%29
t+=+3%2F2seconds. Now substitute this value of t into the original equation and solve for the height, h.
h%283%2F2%29+=+-16%283%2F2%29%5E2%2B48%283%2F2%29
h%283%2F2%29+=+-%2816%289%2F4%29%29+%2B+72
h%283%2F2%29+=+-36%2B72
highlight%28h%283%2F2%29+=+36%29feet. This is the maximum height reached by the ball.
graph%28400%2C400%2C-5%2C5%2C-5%2C40%2C-16x%5E2%2B48x%29