SOLUTION: h= -16t2 +80t + 50 Use this position polynomial to calculate the following: The height of the object after 2 seconds The height of the object after 5 seconds The maximum h

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: h= -16t2 +80t + 50 Use this position polynomial to calculate the following: The height of the object after 2 seconds The height of the object after 5 seconds The maximum h      Log On


   

Question 316746: h= -16t2 +80t + 50
Use this position polynomial to calculate the following:
The height of the object after 2 seconds
The height of the object after 5 seconds
The maximum height of the object
How long the the object will take to reach the ground?
Found 2 solutions by Fombitz, ankor@dixie-net.com:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.h%282%29=+-16%282%29%5E2+%2B80%282%29%2B+50
h%282%29=-64%2B160%2B50
highlight%28+h%282%29=146%29
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2.h%285%29=+-16%285%29%5E2+%2B80%285%29%2B+50
h%285%29=-400%2B400%2B50
highlight%28h%285%29=50%29
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3.Complete the square to get to vertex form.
The max. height occurs at the vertex.
h%28t%29=-16t%5E2%2B80t%2B50
h%28t%29=-16%28t%5E2-5t%2B25%2F4%29%2B50%2B100
h%28t%29=-16%28t-5%2F2%29%5E2%2B150
(5/2,150) is the vertex.
highlight%28hmax=150%29
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4. h%28t%29=-16%28t-5%2F2%29%5E2%2B150
-16%28t-5%2F2%29%5E2%2B150=0
16%28t-5%2F2%29%5E2=150
%28t-5%2F2%29%5E2=75%2F8
t-5%2F2=0+%2B-+sqrt%2875%2F8%29
t=5%2F2+%2B-+%285%2F2%29%2Asqrt%283%2F2%29
The time we need is the positive solution,
t=5%2F2%2B%285%2F2%29%2Asqrt%283%2F2%29
highlight%28t=5%2F2%2B%285%2F4%29%2Asqrt%286%29%29 or approximately
t=5.56 sec.
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.
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+graph%28+300%2C+300%2C+-2%2C+8%2C+-10%2C+190%2C+-16x%5E2+%2B80x+%2B+50%29+

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
h= -16t^2 +80t + 50
;
Use this position polynomial to calculate the following:
:
The height of the object after 2 seconds
h = -16(2^2) + 80(2) + 50
h = -16(4) + 160 + 50
h = -64 + 160 + 50
h = 146 ft
;
The height of the object after 5 seconds
Do the same as above, using t=5
:
The maximum height of the object
Find the axis of symmetry; x = -b/(2a)
in this equation x=t; a=-16: b=80;
t = %28-80%29%2F%282%2A-16%29
t = %28-80%29%2F%28-32%29
t = 2.5 sec; the max height will occur in 2.5 sec
Substitute 2.5 for t to find the max height
h = -16(2.5^2) + 80(2.5) + 50
h = -16(6.25) + 200 + 50
h = -100 + 200 + 50
h = 150 ft is max height
:
How long the the object will take to reach the ground?
when it strikes the ground, h=0, write the equation
-16t^2 + 80t + 50 = 0; solve for t using the quadratic formula
a=-16, b=80, c=50
t+=+%28-80+%2B-+sqrt%2880%5E2-4%2A-16%2A50+%29%29%2F%282%2A-16%29+
I'll let you do the math here, the positive solution ~ 5.5 sec