SOLUTION: 1. A baseball hit straight up in the air is at height h feet above the ground, whre h is given by h(t)= 4+50t-16t², t seconds after being hit. A)State a reasonable domain for th

Question 54474: 1. A baseball hit straight up in the air is at height h feet above the ground, whre h is given by h(t)= 4+50t-16t², t seconds after being hit.
A)State a reasonable domain for this function
B)What is the y -intercept? Interpret the y-intercept relative to this situation
C)When does the baseball hit the ground?
D)When, if ever, is it 30 feet high? 90 feet high?
E)What is the maximum height that the baseball reaches? When does it reach that height?
P.S. I have been sick for like a week and returning to school tom. i have to turn in this worksheet, i was not there so i am freaking out how to do this. I think the y-intercept is 4, would the domain be all real numbers. Please help If you can, thanks
Found 2 solutions by stanbon, Nate:
Answer by stanbon(57984) (Show Source):
You can put this solution on YOUR website!
1. A baseball hit straight up in the air is at height h feet above the ground, whre h is given by h(t)= 4+50t-16t², t seconds after being hit.
A)State a reasonable domain for this function
Maybe 0<=t<=10. The ball is in the air maybe as long as 10 seconds.
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B)What is the y -intercept? Interpret the y-intercept relative to this situation
To find the y-intercept let t=0.
You get h(0)=4+50*0-16*0^2
h(0)=4 ft.
That means the baseball is four feet off the ground when it is first hit.
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C)When does the baseball hit the ground?
It hits the ground when h(t)=0, that is when it's height is zero.
Solve 4+50t-16t^2=0
t=[-50+-sqrt(50^2-4(-16)(4)]/(-32)
t=[-50+-sqrt(2756)]/(-32)
t=3.2 seconds
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D)When, if ever, is it 30 feet high? 90 feet high?
Solve 30=4+50t-16t^2
Solve 90=4+50t-16t^2
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E)What is the maximum height that the baseball reaches? When does it reach that height?
It reaches the maximum when t=-50/(-32)=1.56 seconds
That is approximately half-way between t=0 and t=3.2
Cheers,
Stan H.

Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website!
A)
Well, for a vertical parabola, the domain is always all real numbers (unless acted upon by a different source - not a parabola.) The domain is all real values.
B) Y - Intercept
y = -16(0)^2 + 50(0) + 4
y = 4
(0,4)
C)
0 = -16t^2 + 50t + 4
0 = 8t^2 - 25t - 2

about 3.2 seconds
D)
f(x) = -16t^2 + 50t + 4
v(1.5625,68.56640625)
The highest the ball will go is about 68 feet.
30 = -16t^2 + 50t + 4
0 = -16t^2 + 50t - 26
0 = 8t^2 - 25t + 13

About 2.5 seconds
E)
v(1.5625,68.56640625) Vertex explains it all.
You might have to check if that is right (the vertex).