SOLUTION: Please, please help with this question. I do not know where to go.
When a ball is thrown vertically upward, its height (h, in feet) above the ground after t seconds is describe
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Question 60913: Please, please help with this question. I do not know where to go.
When a ball is thrown vertically upward, its height (h, in feet) above the ground after t seconds is described by the mathematical formula h = - 16t2 + 96t + 80. At what times is the ball 103 feet above the ground?
Found 2 solutions by checkley71, funmath:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
103=-16t^2+96t+80
16t^2-96t-80+103=0
16t^2-96t+23=0
(4x-23)(4x-1)=0
4x-23=0
4x=23
x=23/4
x=5.75 solution 5.75 seconds
4x-1=0
4x=1
x=1/4
x=.25 solution .25 seconds
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
When a ball is thrown vertically upward, its height (h, in feet) above the ground after t seconds is described by the mathematical formula h = - 16t2 + 96t + 80. At what times is the ball 103 feet above the ground?
Use the quadratic formula:
a=16, b=-96, c=23
and
and
and
x=.25 s and x=5.75 s
One the way up, the ball hits 103 ft in .25 s, on the way back down it hits it again in 5.75 s.
Since this came out to rational answers this could have been factored. When the numbers are more difficult to deal with the quadratic formula is easier to use than factoring. Here's what it looks like factored.
4t-1=0 and 4t-23=0
4t=1 and 4t=23
t=1/4 s and t=23/4 s
Happy Calculating!!!
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