This is the answers tab
Firstly, let us analyze the data the problem has given us. This can be achieved through a drawing
Given the equations of a movement with acceleration, we can write them as the following
[tex]S_y(t)=S_{0y}+V_0*sin(32°)*t+\frac{at^2}{2}[/tex][tex]S_x(t)=S_{0x}+V_0*cos(32°)*t[/tex]Given these equations and our data, we can replace some of its unknowns, and we're left with
[tex]S_y(t)=V_0*0.53*t-5t^2[/tex][tex]S_x(t)=V_0*0.848*t[/tex]The first two equations were reduced to these ones by applying the data that the angle is 32°, and that the grenade hit at the same height. Now, we'll plug the information about the distance it reached. This will leave us with the following
[tex]S_x(t_{impact})=V_0*0.848*t_{impact}=39[/tex][tex]S_y(t_{impact})=V_0*0.53*t_{impact}-5*t_{impact}^2[/tex]Then we're left with the following system
[tex]\begin{gathered} 0.53V_0t_{impact}-5t_{impact}^2=0 \\ 0.848V_0t_{impact}=39 \end{gathered}[/tex]On our first equation, we can divide it by t impact, as we know it is not 0
[tex]\begin{gathered} 0.53V_0-5t_{impact}=0 \\ 0.848V_0t_{impact}=39 \end{gathered}[/tex]We can then rearrange and get the following
[tex]\begin{gathered} V_0=\frac{5t_{impact}}{0.53} \\ 0.848V_0t_{impact}=39 \end{gathered}[/tex]By replacing V0 on the lower equation, we get
[tex]0.848*\frac{5t_{impact}}{0.53}*t_{impact}=39[/tex]And this gives us the following equation
[tex]5t_{impact}^2=39*\frac{0.53}{0.848}[/tex]Simplifying...
[tex]t_{impact}^2=4.875[/tex]Finally, applying the square root, we're left with
[tex]t_{impact}\text{ = }2.21s[/tex]So this is the time the Soldier had to escape
