Chapter 3.4, Problem 35E

To determine

To calculate:the exit velocity of lava fountains.

Answer to Problem 35E

Initial velocity of lava fountain that is $v_0$ = $348.71ft/\sec$ which is same as $=237.82\text{miles/hour}$ .

Explanation of Solution

Given information:

1. The height $s$ of the fountain of lava can be given by following formula.

  $s=v_{0}t-16t^2$ where $v_0$ is exit velocity and $t$ is time.

2. The maximum height of lava fountain is $1900$ ft.

Calculation:

Expression representing height of fountain of lava is as follows

  $s=v_{0}t-16t^2$

Differentiate the given expression with respect to $t$ .

  $\frac{ds}{dt}=\frac{d}{dt}\left(v_{0}t-16t^2\right)$

  $\frac{ds}{dt}=v_0\frac{d}{dt}t-16\frac{d}{dt}{t}^2$

  $\frac{ds}{dt}=v_0-32t$

At maximum height $\frac{ds}{dt}=0$

  $0=v_0-32t$

  $v_0=32t$

  $t=\frac{v_0}{32}$

As maximum height is $1900$ ft and $t=\frac{v_0}{32}$ at maximum, substitute $s=1900$ and $t=\frac{v_0}{32}$ to get the value of $v_0$ .

  $1900=v_0\left(\frac{v_0}{32}\right)-16{\left(\frac{v_0}{32}\right)}^2$

  $1900\times 32=v_0{}^2-\frac{16v_0{}^2}{32}$

  $1900\times 32=v_0{}^2-\frac{v_0{}^2}{2}$

  $1900\times 32\times 2=v_0{}^2$

  $v_0=\sqrt{1900\times 32\times 2}$

  $v_0=\sqrt{1900\times 32\times 2}$

  $v_0=348.71ft/\sec$

As $1ft/\sec =0.682\text{miles/hour}$

  $v_0=348.71\times 0.682\text{miles/hour}$

  $=237.82\text{miles/hour}$

Hence initial velocity of lava fountain that is $v_0$ = $348.71ft/\sec$ which is same as $=237.82\text{miles/hour}$ .