# Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 10

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Suppose that a ball is thrown vertically upward with a velocity of $\displaystyle \frac{80\text{ft}}{s}$, then its height after $t$ seconds is $s = 80t - 16t^2$

a.) What is the maximum height reached by the ball?

b.) What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down?

a.) The maximum height can be determined where $\nu(t) = 0$

\begin{equation} \begin{aligned} \nu(t) = s'(t) &= 80 \frac{d}{dt} (t) - 16 \frac{d}{dt} (t^2)\\ \\ s'(t) &= 80(1) - 16 (2t)\\ \\ s'(t) &= 80 - 32t \end{aligned} \end{equation}

when $\nu(t) =0$

\begin{equation} \begin{aligned} 0 &= 80 -32t\\ t &= 2.5 s \end{aligned} \end{equation}

Substituting the value of $t$ to the equation of $s(t)$ we have...

\begin{equation} \begin{aligned} s(2.5) &= 80 (2.5) - 16 (2.5)^2\\ \\ s(2.5) &= 100 \text{ft} \end{aligned} \end{equation}

Therefore, the maximum height the ball can reach is 100 ft.

b.) when $s = 96$ft,

\begin{equation} \begin{aligned} 96 = 80t - 16t^2\\ 16t^2 - 80t + 96 = 0 \end{aligned} \end{equation}

$t = 2$ and $t =3$

Assuming that $t = 2$ represents the time at which the ball is moving upward and $t=3$ is the time at which the ball is moving downward.

The velocity @$t =2s$,

$\displaystyle \nu(t) = 80-32(2) = 16 \frac{\text{ft}}{s}$

The velocity @$t = 3s$,

$\displaystyle \nu(t) = 80 -32 (3) = -16 \frac{\text{ft}}{s}$