SOLUTION: Can you help solve this?
An arrow is shot upward and outward from atop a 81-foot cliff, with an initial velocity of 132 feet per second. The height of the rocket t seconds afte
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Question 139749: Can you help solve this?
An arrow is shot upward and outward from atop a 81-foot cliff, with an initial velocity of 132 feet per second. The height of the rocket t seconds after launch is given by the equation h= -16t^2 + 132t +81. Find the greatest altitude reached by the arrow. Round to the nearest foot.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An arrow is shot upward and outward from atop a 81-foot cliff, with an initial velocity of 132 feet per second. The height of the rocket t seconds after launch is given by the equation h= -16t^2 + 132t +81.
:
The max height will occur at the axis of symmetry
Find the axis of symmetry using the formula: x = -b/(2a);
:
In this equation it would be:
t =
t =
t = +4.125 sec the arrow will be at the greatest altitude
:
Find the greatest altitude reached by the arrow. Round to the nearest foot.
:
Substitute 4.125 for t in the given equation
h = -16(4.125^2) + 132(4.125) + 81
:
h = -16(17.015625) + 544.5 + 81
:
h = -272.25 + 625.5
:
h = 353.25 ~ 353 ft is max height
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