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)
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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