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

Question 173444: The height in feet for a bal thrown upward at 48 feet per second is given by s(t)=-16t^2+48t, where t is the time in seconds after the ball is tossed what is the maximum height that the ball will reach?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If you graph you get a parabola that is concave downward because the coefficient on the term is negative. So, the maximum height will be reached at the time value coordinate of the vertex of the parabola and the maximum height will be the function value at the vertex.

The independent variable coordinate of the vertex of a parabola that is expressed in the form is given by . Here, and , so the time value for the maximum height is

To find the actual maximum height, evaluate Answer: 36 feet.

The trouble with this problem is that you were given a function for the height of a thrown ball that only works in the very unlikely circumstance that the person throwing the ball is standing in a hole whose depth is exactly the height of his/her hand when he/she released the ball. Under ordinary circumstances, a 6-foot tall person would release the ball somewhere in the neighborhood of 7 feet above the ground and therefore your function would be .

Super-Double-Plus Extra Credit.

Assuming the same 7 feet above the ground release point, what would the s(t) function be if the ball were thrown upward at 64 feet per second initial velocity?

I just assumed that you needed the Algebra answer to this question. If you need the Calculus method, write back.
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