The reason we set h equal to 0 is because h is height, and of course the height of an object when it is on the ground is 0. That looks like this then:
![0=-16t^2+1600](https://img.qammunity.org/2019/formulas/mathematics/high-school/plpwhxjwl7btctgfu1y5jnko0vi786pe3s.png)
. To make this easier to factor, aka solving it for the time it takes to hit the ground, we will factor out a -16. That gives us this:
![0=-16(t^2-100)](https://img.qammunity.org/2019/formulas/mathematics/high-school/jfkm90eos5iwsfwpojh1a142etp6cqee37.png)
. That means that, by the Zero Product Property, either -16=0, or
![t^2-100=0](https://img.qammunity.org/2019/formulas/mathematics/high-school/pivg6rd19f8xdaead0al0x1kynaenk7ngp.png)
. Of course
![-16 \\eq 0](https://img.qammunity.org/2019/formulas/mathematics/high-school/lk5uvs242q570azldtnx6ye0091kqt0yds.png)
, so that means that
![t^2-100=0](https://img.qammunity.org/2019/formulas/mathematics/high-school/pivg6rd19f8xdaead0al0x1kynaenk7ngp.png)
. We add 100 to both sides to get
![t^2=100](https://img.qammunity.org/2019/formulas/mathematics/high-school/mqtxkjks81656dne85vap1pu0rtw26qfez.png)
. Taking the square root of both sides, t = -10 and t = 10. The 2 things in math that will never EVER be negative are time and distance/length, so that means that -10 seconds is out. Therefore, t = 10 seconds.