SOLUTION: A ball is thrown upward from the top of a 240 foot building. The ball is 256 feet above ground level a fter 1 second, and it reaches ground level in 5 seconds. the height above gr

Question 724176: A ball is thrown upward from the top of a 240 foot building. The ball is 256 feet above ground level a fter 1 second, and it reaches ground level in 5 seconds. the height above ground is a quadratic function of the time after the ball is thrown. Write the equation of this function.
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A ball is thrown upward from the top of a 240 foot building. The ball is 256 feet above ground level a fter 1 second, and it reaches ground level in 5 seconds. the height above ground is a quadratic function of the time after the ball is thrown. Write the equation of this function.
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It's the common formula for projectiles on Earth.
h(t) = -16t^2 + vt + h, h in feet
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h(1) = -16 + v + 240 = 256
v = 32 ft/sec (the upward speed at launch)
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h(t) = -16t^2 + 32t + 240 is the equation
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h(5) = -400 + 160 + 240 = 0 confirms impact at t = 5 seconds

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The form of the equation is:

given:
ft
When ,
When ,
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(1)
---------------

(2)
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(2)
Add (1) and (2)
(1)
(2)

and, since
(1)
(1)
(1)
The equation is:

Here's the plot:

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