Question

You are part of a search-and-rescue mission that has been called out to look for a lost explorer. You’ve found the missing explorer, but, as FIGURE P11.50 shows, you’re separated from him by a $200-m$-high cliff and a $30-m$-wide-raging river. To save his life, you need to get a 5.0 kg package of emergency supplies across the river. Unfortunately, you can’t throw the package hard enough to make it across. Fortunately, you happen to have a 1.0 kg rocket intended for launching flares. Improvising quickly, you attach a sharpened stick to the front of the rocket so that it will impale itself into the package of supplies, then fire the rocket at ground level toward the supplies. What minimum speed must the rocket have just before impact in order to save the explorer’s life?

Short answer

To preserve the explorer's life, the rocket must have a minimal speed shortly before impact is$28.2m/s$

Step-by-step solution

Step 1. Given information

Cliff height,$y=100m$

Width of the river,$x=30m$

Mass of package,$m_{package}=5kg$

mass of rocket,$m_{rocket}=1kg$

Step 2. Evaluating the velocities

To determine the minimum speed that the system package and rocket must travel to reach the missing explorer, we must first determine the distance traveled:

$$\begin{gathered} y_{\text{final}}=y_{\text{initial}}+v_{y,\text{initial}}\text{Δ}t-\frac{1}{2}g{\left(\text{Δ}t\right)}^2 \\ 0=200m+0-0.5\cdot 9.81m/s^2\cdot {\left(\text{Δ}t\right)}^2 \\ △t=6.38s \end{gathered}$$

We can obtain the quickest speed from there,

$\begin{array}{l}{x}_{\text{final}}=x_{\text{initial}}+v_{x,\text{initial}}\text{Δ}t+\frac{1}{2}a\\ 30s=0+v_{x,\text{initial}}\cdot 6.38s+0\end{array}$

localid="1648099042604" $\begin{array}{rr}\Rightarrow {\mathrm{v}}_{\mathrm{x},\text{initial}}& =4.7\mathrm{m}/\mathrm{s}.\end{array}$

Step 3. Apply the conservation principle

We can calculate the lowest rocket speed required to reach the missing explorer using the law of conservation of momentum:

$\begin{array}{l}{m}_{\text{rocket}}{v}_{\text{rocket}}+m_{\text{package}}{v}_{\text{package}}=\left(m_{\text{rocket}}+m_{\text{package}}\right)v_{\text{system,initial}}\\ 1kg\cdot v_{\text{rocket}}+0=\left(5kg+1kg\right)\cdot 4.7m/s\end{array}$

$\begin{array}{rr}{v}_{\text{rocket}}& =28.2m/s.\end{array}$