Chapter 2, Problem 88RE

Solution

(a)

To find: The equation that models the height of the rocket after t second

The equation that models the height of the rocket after t second is $h=h_0+v_{0}t-\frac{1}{2}g{t}^2$

Given information Initial velocity $v_0=170ft/\sec$ ,initial height $h_0=6ft$

The kinematic equation is $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

Here the motion is upward so $a=-gx=h{x}_0=h_0$

The equation becomes

  $h=h_0+v_{0}t-\frac{1}{2}g{t}^2$

(B)

To find: The maximum height

The maximum height is $h_{\max }=7.61ft$ .

Given information Initial velocity $v_0=170ft/\sec$ , initial height $h_0=6ft$ , $g\approx 32.2ft/s^2$

  $h=h_0+v_{0}t-\frac{1}{2}g{t}^2$

Substitute the given values

  $\begin{array}{l}{h}_{\max }=6+170t-\frac{32.2}{2}{t}^2\\ h(t)=6+170t-16.1t^2\end{array}$

To find the maximum height calculate $h\text{'}(t)=0$

  $\begin{array}{l}170-32.2t=0\\ t=\frac{170}{32.2}\\ t=5.27\end{array}$

Now find $h\text{'}\text{'}(t)$

  $h\text{'}\text{'}(t)=-932.2<0$

Hence the height is maximum at t= 5.27

Substitute t= 5.274 in the $h(t)=6+170t-16.1t^2$

  $\begin{array}{l}h(t)=6+170\cdot \left(5.27\right)-32.2{\left(5.27\right)}^2\\ h_{\max }=7.61ft\end{array}$

(c)

To find: When will rock hit the ground

The rock hit the ground in t=0.034 seconds.

Given information Initial velocity $v_0=170ft/\sec$ ,initial height $h_0=6ft$

  $h(t)=6+170t-4.9t^2$

Rocket will hit the ground then height =0

  $\begin{array}{l}6+170t-4.9t^2=0\\ t=\frac{-170\pm \sqrt{{170}^2-4\cdot 6\cdot \left(-4.9\right)}}{2\left(4.9\right)}\\ t=\frac{-170\pm 170.34}{9.8}\\ t=\frac{-170+170.34}{9.8}\\ t=\frac{0.34}{9.8}=.034\end{array}$

Rocket will hit the ground at t=0.034 seconds