In order to determine the time it takes for the music player to fall to the bottom of the ravine, we shall find the solutions of t as follows;
[tex]\begin{gathered} t=\sqrt[]{\frac{8t+24}{16}} \\ \end{gathered}[/tex]Take the square root of both sides;
[tex]\begin{gathered} t^2=\frac{8t+24}{16} \\ \text{Cross multiply and we'll have;} \\ 16t^2=8t+24 \\ \text{ Re-arrange the terms and we'll now have;} \\ 16t^2-8t-24=0 \end{gathered}[/tex]We can now solve this using the quadratic equation formula;
[tex]\begin{gathered} \text{The variables are;} \\ a=16,b=-8,c=-24 \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-8)\pm\sqrt[]{(-8)^2-4(16)(-24)_{}}}{2(16)} \\ t=\frac{8\pm\sqrt[]{64+1536}}{32} \\ t=\frac{8\pm\sqrt[]{1600}}{32} \\ t=\frac{8\pm40}{32} \\ t=\frac{8+40}{32},t=\frac{8-40}{32} \\ t=\frac{48}{32},t=-\frac{32}{32} \\ t=1.5,t=-1 \end{gathered}[/tex]We shall now plug each root back into the original equation, as follows;
[tex]\begin{gathered} \text{Solution 1:} \\ \text{When t}=1.5 \\ t=\sqrt[]{\frac{8t+24}{16}} \\ t=\sqrt[]{\frac{8(1.5)+24}{16}} \\ t=\sqrt[]{\frac{12+24}{16}} \\ t=\sqrt[]{\frac{36}{16}} \\ t=\frac{6}{4} \\ t=1.5\sec \end{gathered}[/tex][tex]\begin{gathered} \text{Solution 2:} \\ \text{When t}=-1 \\ t=\sqrt[]{\frac{8(-1)+24}{16}} \\ t=\sqrt[]{\frac{-8+24}{16}} \\ t=\sqrt[]{\frac{16}{16}} \\ t=\frac{4}{4} \\ t=1 \end{gathered}[/tex]From the result shown the ballon will deploy after 1.5 seconds for the first solution.
However t = -1 cannot be a solution since you cannot have a negative time (-1 sec)
ANSWER:
t =1.5 is a solution
