SOLUTION: A ball is thrown vertically into the air with an initial velocity at 128 feet per second. It's height ( in feet ) after t second is given by h= 128t-16t^2. when will the ball bet a
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-> SOLUTION: A ball is thrown vertically into the air with an initial velocity at 128 feet per second. It's height ( in feet ) after t second is given by h= 128t-16t^2. when will the ball bet a
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Question 655083: A ball is thrown vertically into the air with an initial velocity at 128 feet per second. It's height ( in feet ) after t second is given by h= 128t-16t^2. when will the ball bet at 192 feet above the ground? Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! A ball is thrown vertically into the air with an initial velocity at 128 feet per second. It's height ( in feet ) after t second is given by h= 128t-16t^2. when will the ball bet at 192 feet above the ground?
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h= 128t-16t^2
192=128t-16t^2
-16t^2+128t-192=0
This is an equation of a parabola that opens downwards
simplify, divide by 16
-t^2+8-12=0
t^2-8+12=0
(t-6)(t-2)=0
t=6
or
t=2
The ball will be at 192 feet above the ground:
after 2 seconds on the way up
after 6 seconds on the way down
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