A ball is thrown upward. its height (h, in feet) is modeled by the function h = - 16t^2 + 64t+3, where t is the length of time (in seconds) that the ball has been in the air. what is the maximum height the ball reaches? a ball is thrown upward. its height (h, in feet) is modeled by the function h = - 16t^2 + 64t+3, where t is the length of time (in seconds) that the ball has been in the air. what is the maximum height the ball reaches?

Question

Kolton Hickman

Mathematics

2 May, 05:29

A ball is thrown upward. its height (h, in feet) is modeled by the function h = - 16t^2 + 64t+3, where t is the length of time (in seconds) that the ball has been in the air. what is the maximum height the ball reaches? a ball is thrown upward. its height (h, in feet) is modeled by the function h = - 16t^2 + 64t+3, where t is the length of time (in seconds) that the ball has been in the air. what is the maximum height the ball reaches?

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Zander Farmer · Answer 1

To find the maximum height just simply find the vertex of - 16t^2 + 64t + 3 and to find the axis of symmetry or the x value of the vertex do - b/2a or - 64/-32 = 2 in this situation. Plug in to get the y value - 16 (2) ^2 + 64 (2) + 3 = 67 Vertex (2,67) So the max height is 67 feet (takes 2 seconds to do so)