SOLUTION: If a rock is thrown upward at a speed of 64 ft/sec and over the edge of a cliff 80 feet above a river, its heights h after t seconds is given by the equation {{{h=80+64t-16t^2}}}.
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-> SOLUTION: If a rock is thrown upward at a speed of 64 ft/sec and over the edge of a cliff 80 feet above a river, its heights h after t seconds is given by the equation {{{h=80+64t-16t^2}}}.
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Question 229227: If a rock is thrown upward at a speed of 64 ft/sec and over the edge of a cliff 80 feet above a river, its heights h after t seconds is given by the equation . After how many seconds will it hit the water (h=0)? Solve by factoring the polynomial, using the Zero Factor Property, then solving the linear equations. Answer by ankor@dixie-net.com(22740) (Show Source):
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After how many seconds will it hit the water (h=0)?
: = 0
normal form = 0
Factor out -16, simplifies and changes the signs
-16(t^2 - 4t - 5) = 0
Factors to
-16(t - 5)(t + 1) = 0
The positive solution
t - 5 = 0
t = +5 sec to hit the water
;
:
prove this to yourself
h = 80 + 320 - 16(25)
h = 400 - 400
h = 0
