Question

When does ball 1 reach the ground? round to the nearest tenth.

Answer 1

Given the function:

`\begin{gathered} f(t)=-16t^2+h \\ where \\ f(t)\Rightarrow current\text{ height} \\ t\Rightarrow time \\ h\Rightarrow height \end{gathered}`

When ball 1 reaches the ground, we have

`f(t)=0`

By substitution, we have

`\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} \\ (-h)/(-16)=(-16t^2)/(-16) \\ \Rightarrow t^2=(h)/(16) \\ but\text{ h = 117 for ball 1} \\ thus, \\ t^2=(117)/(16) \\ take\text{ the square root of both sides} \\ √(t^2)=\sqrt{(117)/(16)} \\ \Rightarrow t=\pm2.70416 \\ but\text{ t cannot be negative.} \\ \therefore \\ t=2.7\text{ \lparen nearest tenth\rparen} \end{gathered}`

Hence, ball 1 will reach the ground at

`t=2.7`