Answer:
The first listed equation is the correct equation to solve
The correct answer is: t = 5 seconds
Explanation:
Let me mention first that there is an error in the statement that the gravitational pull of the Earth is 16 ft/s^2. It is in fact 32 ft/s^2, and the actual equation uses half of the acceleration multiplied by the square of the variable time, so it gives as final expression :
![y(t)=-16\,t^2+ 0 \,t +400](https://img.qammunity.org/2021/formulas/mathematics/high-school/y2h6nfmae9f7nh1qw2aws6otnm2gg99atr.png)
and we want to find the value/s for "t" that make this equation equal zero (when it reaches the ground and the object just touches the ground. This makes the equation we want to solve:
![0=-16\,t^2+ 0 \,t +400\\16\,t^2+0\,t -400=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/8273kgri1sktajgy00k9jzxjh1nwgmqefr.png)
which solving for "t" becomes:
![16\,t^2-400 =0\\16\,t^2=400\\t^2=(400)/(16) \\t^2=25\\t= +/- 5\,\,sec](https://img.qammunity.org/2021/formulas/mathematics/high-school/6f3ijauijbvrolcesx2v1v0gxswdgflzl0.png)
So we adopt the positive answer (positive time) since the negative value has no physical meaning for this problem.
That is: t = 5 seconds