Asked by Anonymous
A baseball is thrown into the air with an upward velocity of 25 ft./s. It's height in feet after T seconds can be modeled by the function h(t)=-16t^2 +25t+5. how long will it take the ball to reach it's maximum height? What is the balls maximum height?
Answers
Answered by
Damon
how about complete the square to find the vertex?
16 t^2 -25 t = -h + 5
t^2 -1.5625 t = -(1/16)h + .3125
(t - .78125)^2 = -(1/16 )h + .3125 +.61035
(t - .78125)^2 = -(1/16 )h + .923
(t - .78125)^2 = -(1/16 )(h - 14.8)
t = .781
h = 14.8
16 t^2 -25 t = -h + 5
t^2 -1.5625 t = -(1/16)h + .3125
(t - .78125)^2 = -(1/16 )h + .3125 +.61035
(t - .78125)^2 = -(1/16 )h + .923
(t - .78125)^2 = -(1/16 )(h - 14.8)
t = .781
h = 14.8
Answered by
Damon
or
0 = 25 - 32 t
t = .781
h = 5 + 25(.781) - 16(.781)^2 = 14.8
0 = 25 - 32 t
t = .781
h = 5 + 25(.781) - 16(.781)^2 = 14.8
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