SOLUTION: Hi, I am having trouble figuring out this problem. Would someone please help me figure out the formula? Thank you. The height in feet for a ball thrown upward at 48 feet per s

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Question 197762: Hi, I am having trouble figuring out this problem. Would someone please help me figure out the formula? Thank you.
The height in feet for a ball 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?
Equation of motion of the ball is s%28t%29=+-16t%5E%282%29+%2B+48t
It is the equation of a parabola.
So the maximum height will be at its vertex.
Vertex = (%28-b%2F2a%29,%28s%28-b%2F2a%29%29)
f%28t%29=ax%5E2%2Bbx%2Bc-------(1)
s%28t%29=-16t%5E2%2B48t--------(2)
Comparing (1) and (2)
a=-16
b=48
-b%2F2a=-48%2F%282%2A-16%29= 1.5
s%281.5%29=-16%281.5%29%5E2%2B48%281.5%29=36
So the maximum height the ball will reach is 36 feet

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