SOLUTION: the height in feet for a ball thrown upward at 48 feet persecond is given by s(t)=-16t2+48t, where t is the time in seconds after the ball is tossed. what is the maximum height t

Question 10369: the height in feet for a ball thrown upward at 48 feet persecond is given by s(t)=-16t2+48t, where t is the time in seconds after the ball is tossed. what is the maximum height that the ball will reach?

Please help me with this problem...
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
This is the equation for a parabola that opens downward. The quadratic equation is already in standard form: where x = t, a = -16, b = 48, and c = 0.
The maximum height of the thrown object will be at the vertex (the maximumum value) (h, t) of the parabola.
The t-coordinate (equivalent to the x-coordnate) of the vertex (this is the time at which the object reaches its maximum height) is given by where a = -16 and b = 48.
so, at t= 3/2 secs, the object is at its maximum height.
To find the value of the maximum height at t = 3/2 secs, substitute t = 3/2 into the original quadratic equation and solve for h.

As a check, you could take the first derivative of the quadratic equation and set it to zero to find the value of t at the maximum height.

= secs, same as previous answer for the time the object reaches maximum height.

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