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It got cut off in the picture, but the full question is: A ball is thrown upward into the air so that its height, h, in feet, after t seconds is modeled by the function h(t) = 56t – 16t^2. What is the velocity, in feet per second, of the ball after 2 seconds?

It got cut off in the picture but the full question is A ball is thrown upward into the air so that its height h in feet after t seconds is modeled by the funct class=

Respuesta :

the velocity is the derivative of the position, then we need have to find the derivative:

[tex]v=\frac{dh}{dt}=56-32t[/tex]

Now that we have the velocity in any given time we just evaluate it at t=2:

[tex]\begin{gathered} v=56-32(2) \\ v=56-64 \\ v=-8 \end{gathered}[/tex]

Therefore the velocity is -8 ft/s

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