Question 139749
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. 
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The max height will occur at the axis of symmetry
Find the axis of symmetry using the formula: x = -b/(2a); 
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In this equation it would be:
t = {{{(-132)/(2*-16)}}}
t = {{{(-132)/(-32)}}}
t = +4.125 sec the arrow will be at the greatest altitude
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Find the greatest altitude reached by the arrow. Round to the nearest foot.
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Substitute 4.125 for t in the given equation
h = -16(4.125^2) + 132(4.125) + 81
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h = -16(17.015625) + 544.5 + 81
:
h = -272.25 + 625.5
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h = 353.25 ~ 353 ft is max height