SOLUTION: a baseball is hit straight upward with the initial velocity of 80 ft/sec from a height of 4 ft. The height of the baseball above the ground t sec after it is hit is modeled by the
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-> SOLUTION: a baseball is hit straight upward with the initial velocity of 80 ft/sec from a height of 4 ft. The height of the baseball above the ground t sec after it is hit is modeled by the
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Question 733210: a baseball is hit straight upward with the initial velocity of 80 ft/sec from a height of 4 ft. The height of the baseball above the ground t sec after it is hit is modeled by the function h(t)=-16t^2 + 80t + 4, for what values of t is the baseball 40 ft above the ground? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a baseball is hit straight upward with the initial velocity of 80 ft/sec from a height of 4 ft. The height of the baseball above the ground t sec after it is hit is modeled by the function h(t)=-16t^2 + 80t + 4, for what values of t is the baseball 40 ft above the ground?
:
h(t) = 40, therefore we have the equation
-16t^2 + 80t + 4 = 40
-16t^2 + 80t + 4 - 40 = 0
-16t^2 + 80t - 36 = 0
Easier to factor if you simplify and change the signs, divide by -4, results:
4t^2 - 20t + 9 = 0
Factors to
(2t - 9)(2t - 1) = 0
2t = 1
t = .5 sec it's at 40 ft on the way up
and
2t = 9
t = 4.5 sec its at 40 ft on the way down
:
Graphically, green hor line is at 40'
