Solution:
Given the function:
[tex]\begin{gathered} f(t)=-16t^2+h \\ where \\ f(t)\Rightarrow current\text{ height} \\ t\Rightarrow time \\ h\Rightarrow height \end{gathered}[/tex]
When ball 1 reaches the ground, we have
[tex]f(t)=0[/tex]
By substitution, we have
[tex]\begin{gathered} 0=-16t^2+h \\ add\text{ -h to both side of the equation} \\ 0-h=-16t^2+h-h \\ \Rightarrow-h=-16t^2 \\ divide\text{ both sides by -16} \\ \frac{-h}{-16}=\frac{-16t^2}{-16} \\ \Rightarrow t^2=\frac{h}{16} \\ but\text{ h = 117 for ball 1} \\ thus, \\ t^2=\frac{117}{16} \\ take\text{ the square root of both sides} \\ \sqrt{t^2}=\sqrt{\frac{117}{16}} \\ \Rightarrow t=\pm2.70416 \\ but\text{ t cannot be negative.} \\ \therefore \\ t=2.7\text{ \lparen nearest tenth\rparen} \end{gathered}[/tex]
Hence, ball 1 will reach the ground at
[tex]t=2.7[/tex]
