Solution:
Given the function:
[tex]\begin{gathered} f(t)=-16t^2+h---\text{ equation 1} \\ where \\ f(t)\Rightarrow current\text{v height} \\ t\Rightarrow time \\ h\Rightarrow height \end{gathered}[/tex]
When ball 2 reaches the ground, we have
[tex]f(t)=0----\text{ equation 2}[/tex]
By substitution, we have
[tex]\begin{gathered} 0=-16t^2+h \\ add\text{ -h to both sides of the equation} \\ 0-h=-16t^2+h-h \\ \Rightarrow-h=-16t^2 \\ \therefore \\ t^2=\frac{h}{16}---\text{ equation 3} \end{gathered}[/tex]
To find the time ball 2 reaches the ground, we have
[tex]\begin{gathered} From\text{ equation 3,} \\ t^2=\frac{h}{16} \\ but\text{ h = 277} \\ t^2=\frac{277}{16} \\ take\text{ the square root of both sides} \\ \sqrt{t^2}=\sqrt{\frac{277}{16}} \\ t=\pm4.16082 \\ but\text{ t cannot be negative.} \\ \therefore \\ t\approx4.16\text{ \lparen nearest hundredth\rparen} \end{gathered}[/tex]
Hence, ball 2 reaches the ground at
[tex]t=4.16[/tex]
