Answer:
The first listed equation is the correct equation to solve
The correct answer is: t = 5 seconds
Step-by-step 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 :
[tex]y(t)=-16\,t^2+ 0 \,t +400[/tex]
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:
[tex]0=-16\,t^2+ 0 \,t +400\\16\,t^2+0\,t -400=0[/tex]
which solving for "t" becomes:
[tex]16\,t^2-400 =0\\16\,t^2=400\\t^2=\frac{400}{16} \\t^2=25\\t= +/- 5\,\,sec[/tex]
So we adopt the positive answer (positive time) since the negative value has no physical meaning for this problem.
That is: t = 5 seconds
