To answer this question we will set and solve an equation.
Since the height of the ball above the ground after t seconds is:
[tex]H(t)=(97t-16t^2)ft,[/tex]
then its velocity after t seconds is:
[tex]H^{\prime}(t)=(97-32t)\frac{ft}{s}.[/tex]
Setting H'(t)=48.5ft/s we get:
[tex]48.5\frac{ft}{s}=(97-32t)\frac{ft}{s}.[/tex]
Therefore:
[tex]48.5=97-32t.[/tex]
Subtracting 97 from the above equation we get:
[tex]\begin{gathered} 48.5-97=97-32t-97, \\ -48.5=-32t. \end{gathered}[/tex]
Dividing the above equation by -32 we get:
[tex]\begin{gathered} \frac{-48.5}{-32}=\frac{-32t}{-32}, \\ t\approx1.516. \end{gathered}[/tex]
Answer:
[tex]t=1.516\text{ seconds}[/tex]
