SOLUTION: A ball is thrown straight upward at an initial speed of 104 feet per seconds. From Physics it is known that, after t seconds, the ball reaches a height h feet given by the formula

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Question 398198: A ball is thrown straight upward at an initial speed of 104 feet per seconds. From Physics it is known that, after t seconds, the ball reaches a height h feet given by the formula h=-16t^2 + 104 t.
1.The ball reaches the height of 64.96 feet after t1= _____ and t2=_____ seconds (with t1 < t2)
I set up the problem as 64.96= -16t^2+104t. I solved for t and I keep coming up with t1= .777 and t2= 5.222. What am I doing wrong?
Thank you!

Found 2 solutions by scott8148, stanbon:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
why do you think that your answers are incorrect?

the ball passes through the given height on the way up and on the way down

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A ball is thrown straight upward at an initial speed of 104 feet per seconds. From Physics it is known that, after t seconds, the ball reaches a height h feet given by the formula h=-16t^2 + 104 t.
1.The ball reaches the height of 64.96 feet after t1= _____ and t2=_____ seconds (with t1 < t2)
Solve: -16t^2+104t-64.96 = 0
t = [-104 +- sqrt(104^2 -4*-16*-64.96)]/(2*-16)
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t = [-104 +- sqrt(6658)]/-32
t = [-104 +- 81.6]/(-32)
t = 5.8 seconds or to = 0.7 seconds
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Cheers,
Stan H.
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