Question

When does ball 2 reach the ground? round to the nearest hundredth.

Answer 1

Given the function:

`\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}`

When ball 2 reaches the ground, we have

`f(t)=0----\text{ equation 2}`

By substitution, we have

`\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=(h)/(16)---\text{ equation 3} \end{gathered}`

To find the time ball 2 reaches the ground, we have

`\begin{gathered} From\text{ equation 3,} \\ t^2=(h)/(16) \\ but\text{ h = 277} \\ t^2=(277)/(16) \\ take\text{ the square root of both sides} \\ √(t^2)=\sqrt{(277)/(16)} \\ t=\pm4.16082 \\ but\text{ t cannot be negative.} \\ \therefore \\ t\approx4.16\text{ \lparen nearest hundredth\rparen} \end{gathered}`

Hence, ball 2 reaches the ground at

`t=4.16`