SOLUTION: A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
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Question 652686: A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
a) How long does it take the ball to reach a height of 18ft?
b) How long does it take the ball to hit the ground?
Found 2 solutions by solver91311, stanbon:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
a. Substitute the height value 18 for the variable representing height, 
, then put the equation into standard form and solve the factorable quadratic. You will have two values, one is the time it will reach 18 feet on the way up and the other is the time it will be at 18 feet a second time on the way down.
b. Substitute the height value 0 for the variable representing height (the height of the ground is zero), , then solve the factorable quadratic. Discard the negative root because you don't care what happened before he threw the ball. (Do you?!)
Check Stanbon's (above) arithmetic very carefully. Both of these quadratics are factorable over the Rationals.
John
My calculator said it, I believe it, that settles it
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
a) How long does it take the ball to reach a height of 18ft?
-----
Solve:
-16t^2 + 30t + 4 = 18
----
-16t^2 + 30t - 14 = 0
----
-8t^2 + 15t - 7 = 0
-----
t = [-15 +- sqrt(225 - 4*-8*-7)]/(-16)
----
t = [-15 +- sqrt(29)]/-16
---
t = 0.6 seconds and t = 1.274 seconds
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b) How long does it take the ball to hit the ground?
Solve: -16t^2 + 30t - 4 = 0
-8t^2 + 15t - 2 = 0
t = 1.73 seconds
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Cheers,
Stan H.
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