SOLUTION: If a ball is hit upward at 80 ft per second, then its approximate height, in meters, t seconds later is given by h(t) = 80t − 16t2. When does the ball hit the ground?

Question 509301: If a ball is hit upward at 80 ft per second, then its approximate height, in meters, t seconds later is given by h(t) = 80t − 16t2. When does the ball hit the ground?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If a ball is hit upward at 80 ft per second, then its approximate height, in meters, t seconds later is given by h(t) = 80t − 16t2. When does the ball hit the ground?
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That's the height in feet, not meters.
It hits the ground when h = 0
h(t) = 80t - 16t^2 = 0
80 = 16t
t = 5 seconds
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h(t) = 0 at t = 0 also, when the ball is "launched"

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SOLUTION: If a ball is hit upward at 80 ft per second, then its approximate height, in meters, t seconds later is given by h(t) = 80t &#8722; 16t2. When does the ball hit the ground?

SOLUTION: If a ball is hit upward at 80 ft per second, then its approximate height, in meters, t seconds later is given by h(t) = 80t − 16t2. When does the ball hit the ground?