SOLUTION: Please help me solve this question.
A ball is thrown vertically upward. After t seconds, it's height h (in feet) is given by the function h(t)= 104t-16t^2. What is the maximum he
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A ball is thrown vertically upward. After t seconds, it's height h (in feet) is given by the function h(t)= 104t-16t^2. What is the maximum he
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Question 375307: Please help me solve this question.
A ball is thrown vertically upward. After t seconds, it's height h (in feet) is given by the function h(t)= 104t-16t^2. What is the maximum height that the ball will reach? Do not round your answer. Height: ___ ft Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! h(t)= 104t-16t^2
max height is at the vertex.
.
Vertex is when:
t = -b/(2a) = -104/(2*(-16)) = -104/(-32) = 3.25
.
h(3.25)= 104(3.25)-16(3.25)^2
h(3.25)= 338-169
h(3.25)= 169 feet