Given
![h=160-16t^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/ic9116ezvkah1ge2aan25juy98940487as.png)
Where h is the height of the falling rock at time t, set h=0 and solve for t, as shown below
![\begin{gathered} 160-16t^2=0 \\ \Rightarrow16t^2=160 \\ \Rightarrow t^2=10 \\ \Rightarrow t=\pm√(10) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/4dbbu70jz1jvpuhg63mwqatw4y5oi3cnrs.png)
However, it is not possible that t<0 as it would imply that the rock reached the ground before falling; therefore, the only possible answer is t=sqrt(10).
Rounding the answer,
![\begin{gathered} \Rightarrow t=√(10) \\ \Rightarrow t\approx3.2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/m91geg4fgm0cclgcv585sacgs9l8em1gad.png)
Hence, the answer is t=3.2