Chapter 4.2, Problem 50PPS

a.

To determine

To find: the time at which the water to hit the nozzle on the way down.

Answer to Problem 50PPS

  $t=1.25\sec$ .

Explanation of Solution

Given:

A water hose can spray water at an initial velocity of $40\text{ ft/sec}$ .

Calculation:

According to the question $h\left(t\right)=v_{0}t-16t^2$ .

Where t is the time, and $h\left(t\right)$ is the height of water $v_0\left(t\right)$ is initial velocity.

  $\begin{array}{l}h\left(t\right)=v_{0}t-16t^2\left\{\text{given equation}\right\}\\ h\left(t\right)=40-16t^2\left\{\text{substitutes these value}\right\}\\ \therefore t=\frac{-40}{2\left(-16\right)}\\ t=\frac{5}{4}=1.25\sec \left\{\text{divide both side by 16}\right\}\\ t^2=3.75\\ t=1.94\sec \{\text{taking square root}}\end{array}$

Hence, the time at which the water to hit the nozzle on the way down is $1.25\sec$

b.

To determine

To find: the maximum height of the water, if the height of nozzle is 5 feet above the ground.

Answer to Problem 50PPS

  $h{\left(t\right)}_{\max }=30ft$ .

Explanation of Solution

Given:

A water hose can spray water at an initial velocity of $40\text{ ft/sec}$ .

Calculation:

According to the question $h\left(t\right)=v_{0}t-16t^2$ .

Where t is the time, and $h\left(t\right)$ is the height of water $v_0\left(t\right)$ is initial velocity.

From part (a) $t=1.25\sec$

  $\begin{array}{l}h\left(t\right)=v_{0}t-16t^2\left\{\text{given equation}\right\}\\ h\left(t\right)=40-16{\left(1.25\right)}^2\left\{\text{substitutes these value}\right\}\\ \therefore h\left(t\right)=25ft\\ \{\text{since, the height of the nozzle is 5 ft above the ground}}\\ h{\left(t\right)}_{\max }=30ft\end{array}$

Hence, the maximum height of the water is $h{\left(t\right)}_{\max }=30ft$ .