Question

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?

Answer 1

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

`v=(dh)/(dt)=56-32t`

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

`\begin{gathered} v=56-32(2) \\ v=56-64 \\ v=-8 \end{gathered}`

Therefore the velocity is -8 ft/s