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A ball is thrown into the air with an upward velocity of 44 ft/s. Its height h in feet after t seconds is given by the function [tex]h=-16t ^{2} +44t+9[/tex]. What is the height of the ball after 2 seconds?



Answer :

[tex]h=-16t^2+44t+9\\\\substitute\ t=2:\\\\h=-16\cdot2^2+44\cdot2+9=-16\cdot4+88+9=-64+97=33\\\\Answer:33\ feets.[/tex]
kate200468
[tex]the\ function:\ \ h(t)=-16t^2+44t+9\\ \\ h=2\ \ \ \Rightarrow\ \ \ h(2)=-16\cdot 2^2+44\cdot 2+9=-16\cdot4+88+9=33\\ \\Ans.\ the\ height\ of\ the\ ball\ after\ 2\ seconds\ is\ 33\ feets[/tex]

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