SOLUTION: The height in feet for a ball thrown upward at 48 feet per second is given by
s(t)= -16t^2 + 48feet, where t is the time in seconds after the ball is tossed. What is the maximum
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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 + 48feet, where t is the time in seconds after the ball is tossed. What is the maximum
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Question 225644: The height in feet for a ball thrown upward at 48 feet per second is given by
s(t)= -16t^2 + 48feet, 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! s(t)= -16t^2 + 48t
Find the 2 times that s = 0. It takes the same time to go up as to fall back down, so at the mid-time it's at max height.
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-16t^2 + 48t = 0
t = 0 (at the start)
t = 3 (return to Earth)
t at max height = 1.5 seconds
s(1.5) = -16*2.25 + 48*1.5
= -36 + 72
= 36 ft
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Or, find the vertex of the parabola
It's at t = -b/2a = -48/-32 = 1.5
Same answer.
