Chapter 9.4, Problem 38PPE

a)

To determine

To evaluate: the vertical motion model to write an equation that gives the ball’s height h , in feet, at time t , in seconds.

Answer to Problem 38PPE

Equation becomes $h=-16t^2+38t+5$ .

Explanation of Solution

Given:

Suppose a softball thrown into the air with an initial upward velocity of 38 ft/s and an initial height of 5ft .

The vertical model is given by,

  $h=-16t^2+vt+s$

where

h = total height reached,

v = velocity

s = initial height

After substitution,

Equation becomes $h=-16t^2+38t+5$ .

b)

To determine

To evaluate: the equation written in part (a) for h = 0 to find when the ball lands.

Answer to Problem 38PPE

The ball lands when t = 2.5 sec.

Explanation of Solution

Given:

Suppose a softball thrown into the air with an initial upward velocity of 38 ft/s and an initial height of 5ft .

The vertical model is given by,

  $h=-16t^2+vt+s$

where,

h = total height reached,

v = velocity

s = initial height

After substitution,

Equation becomes $h=-16t^2+38t+5$ .

When h = 0,

  $-16t^2+38t+5=0$

Applying quadratic formula to find the value of ‘ t’

  $\begin{array}{c}t=\frac{-38\pm \sqrt{{\left(38\right)}^2-4\left(-16\right)\left(5\right)}}{2\left(-16\right)}\\ =\frac{-38\pm \sqrt{1444+320}}{-32}\\ =\frac{-38\pm \sqrt{1764}}{-32}\\ =\frac{-38\pm 42}{-32}\\ =\frac{-80}{-32},-\frac{4}{32}\\ =2.5,-0.125\end{array}$

The negative value of t is not possible.

So, the ball lands when t = 2.5 sec.