SOLUTION: h(t)=-2.7t(squared)+v0t+h0, v0=40 ft/sec. h0=5ft. t =the number of sec. after the ball is thrown. need to know how to find any coordinate. how many secs is the ball 25 ft. after 5

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: h(t)=-2.7t(squared)+v0t+h0, v0=40 ft/sec. h0=5ft. t =the number of sec. after the ball is thrown. need to know how to find any coordinate. how many secs is the ball 25 ft. after 5       Log On


   

Question 287936: h(t)=-2.7t(squared)+v0t+h0, v0=40 ft/sec. h0=5ft. t =the number of sec. after the ball is thrown. need to know how to find any coordinate. how many secs is the ball 25 ft. after 5 secs what is the height of the ball ect.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
h(t)=-2.7t(squared)+v0t+h0, v0=40 ft/sec. h0=5ft. t =the number of sec. after the ball is thrown. need to know how to find any coordinate. how many secs is the ball 25 ft. after 5 secs what is the height of the ball ect.
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h(t) = -2.7t^2 + vot + ho
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Substitute the given information:
h(t) = -2.7t^2 + 40t + 5
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Solve -2.7t^2 + 40t+5 = 25
-2.7t^2 + 40t - 20 = 0
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t = [-40 +- sqrt(40^2-4*-2.7*-20)]/(2*-2.7)
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t = [-40 +- 37.20]/(-5.4)
0.518<= t <=14.3
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graph%28400%2C300%2C-5%2C15%2C-5%2C80%2C25%2C-2.7x%5E2%2B40x-20%29
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Cheers,
Stan H.