Question

I have a calculus question about derivatives as rates of change change with a ball thrown in the air and its velocity picture included

Answer 1

To answer this question we will set and solve an equation.

Since the height of the ball above the ground after t seconds is:

`H(t)=(97t-16t^2)ft,`

then its velocity after t seconds is:

`H^(\prime)(t)=(97-32t)(ft)/(s).`

Setting H'(t)=48.5ft/s we get:

`48.5(ft)/(s)=(97-32t)(ft)/(s).`

Therefore:

`48.5=97-32t.`

Subtracting 97 from the above equation we get:

`\begin{gathered} 48.5-97=97-32t-97, \\ -48.5=-32t. \end{gathered}`

Dividing the above equation by -32 we get:

`\begin{gathered} (-48.5)/(-32)=(-32t)/(-32), \\ t\approx1.516. \end{gathered}`

Answer:

`t=1.516\text{ seconds}`