SOLUTION: The height of a ball thrown straight up into the air can be modeled by
h(t) = −16t2 + 30t + 3.5
where h(t) is the height of the ball in feet t seconds after it is thrown.
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Quadratic Equations and Parabolas
-> SOLUTION: The height of a ball thrown straight up into the air can be modeled by
h(t) = −16t2 + 30t + 3.5
where h(t) is the height of the ball in feet t seconds after it is thrown.
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Question 1025687: The height of a ball thrown straight up into the air can be modeled by
h(t) = −16t2 + 30t + 3.5
where h(t) is the height of the ball in feet t seconds after it is thrown.
When will the ball be at a height of 15 feet? (Round your answers to two decimal places. Include units with your numerical answers.)
The ball is at a height of 15 feet_______(smaller value) after being thrown and again_______(larger value) after being thrown. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! h(t) = −16t^2 + 30t + 3.5
15=-16t^2+30t+3.5
16t^2-30t+11.5=0
t=(1/32)(30 +/- sqrt (900-736)); sqrt (164)=12.81)
t=(1/32)(30+/-12.81)
t=(1/32)(42.81) and (1/32)(17.19)
t=1.34 sec and t=0.537 sec
in order at 0.537 and 1.34 sec.
