Question

The height of a ball thrown vertically upward is given by the function h(t)=−16t²+40t+6, where h is in feet and t is in seconds. What is the speed of the ball at t = 1 in ft./s?

Answer 1

Write out the function foir the height

`h(t)=-16t^2+40t+6`

The speed of an object is the rate of change of distance with time. Since rate of change is also differentiation, hence we diferentiate the function above

Applying the rule of differentiation, we have

Recall the rule for diffentiation

`\begin{gathered} \text{If } \\ y=x^n \\ \text{Then} \\ (dy)/(dx)=ny^(n-1) \end{gathered}`

hence, we have

`\begin{gathered} (dh)/(dt)=-16*2t^(2-1)+40*1t^(1-1) \\ \\ (dh)/(dt)=-32t+40 \end{gathered}`

Then the speed of the ball is goven by the function

`V=(dh)/(dt)=-32t+40`

At t= 1, we substitute into the function we obtained for speed

Hence, we have

`\begin{gathered} V=-32(1)+40 \\ Where\text{ t=1} \\ V=-32+40 \\ V=8\text{fts}^(-1) \end{gathered}`

hence

The speed of the ball at t=1s is 8ft/s