SOLUTION: 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 th
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-> SOLUTION: 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 th
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Question 173473: 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? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 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?
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I don't know what level math class you have, but to find a max you set the 1st derivative to zero.
s(t)=-16t^2+48t
ds/dt = -32t + 48 = 0
32t = 48
t = 1.5 seconds at the max height
s(1.5) = 16*2.25
= 36 feet
