Answer:
Time to land on 11ft is 4.25 seconds
Explanation:
Given
![y = 300 - 16t^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/d0ht0mhw0b912gz1spqyppuwhkuc6vjyg9.png)
Required
Solve for t when height is 11ft
This implies that
![y = 11](https://img.qammunity.org/2021/formulas/mathematics/college/op8nhqwf4be0epmzl53hwwems5evyf6qpl.png)
So: substitute 11 for y in
![y = 300 - 16t^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/d0ht0mhw0b912gz1spqyppuwhkuc6vjyg9.png)
![11 = 300 - 16t^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/bt1j7am448u3us9dfvtcyvrqb19f98w49p.png)
Collect Like Terms
![16t^2 = 300 - 11](https://img.qammunity.org/2021/formulas/mathematics/high-school/6gkipecs8ip68ttigodldopj21i8gotvlw.png)
![16t^2 = 289](https://img.qammunity.org/2021/formulas/mathematics/high-school/aufu6rcyor33ivhn7ay5lam86x794vxez6.png)
Express both sides as squares
![4^2 * t^2 = 17^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/3ahc4r3kfksm66rut2wb8cu4j732mp3ukf.png)
![(4t)^2 = 17^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/rxjsei28u9m92owc2yafedzbng1mnt0vva.png)
Take square roots of both sides (ignore negative)
![4t = 17](https://img.qammunity.org/2021/formulas/mathematics/high-school/4mdmm8uxu2sontusq5xk3vsy77kh944cli.png)
Solve for t
![t = (17)/(4)](https://img.qammunity.org/2021/formulas/mathematics/high-school/xv29yomvea66lopoqehfa74gbu4h9d4080.png)
![t = 4.25](https://img.qammunity.org/2021/formulas/mathematics/high-school/9c89j8zi2x3rp19nxyygjr4azahv1djuax.png)
Hence, time to land on 11ft is 4.25 seconds