When the ball hits the ground, the height of the ball to the ground is h(t) = 0.
Therefore, we can now substitute h(t), and solve the function using quadratic formula.
[tex]\begin{gathered} h(t)=-16t^2+16t+400 \\ 0=-16t^2+16t+400 \\ \\ \text{The function is now in standard form where} \\ a=-16,b=16,c=400 \end{gathered}[/tex]
Using the quadratic formula, substitute the following values a,b, and c.
[tex]\begin{gathered} t=\frac{ -b \pm\sqrt{b^2 - 4ac}}{ 2a } \\ t=\frac{ -16 \pm\sqrt{16^2 - 4(-16)(400)}}{ 2(-16) } \\ t=\frac{-16\pm\sqrt[]{256-(-25600)}}{-32} \\ t=\frac{ -16 \pm\sqrt{25856}}{ -32 } \\ t=\frac{ -16 \pm16\sqrt{101}\, }{ -32 } \\ \\ t=\frac{-16+16\sqrt{101}}{-32} \\ t=-4.52494 \\ \\ t=\frac{-16-16\sqrt[]{101}\, }{-32} \\ t=5.52494 \end{gathered}[/tex]
We have two solutions, t = -4.52494, and t = 5.52494. However, we will disregard the negative time value.
Therefore, the ball will hit the ground after 5.52494 seconds.
