Answer :
Here's our equation.
[tex]h=-16t+64t+3[/tex]
We want to find out when it returns to ground level (h = 0)
To find this out, we can plug in 0 and solve for t.
[tex]0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}[/tex]
[tex]= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}[/tex]
So the ball will return to the ground at the positive value of [tex]\boxed{\frac{8+\sqrt{67}}{4}}[/tex] seconds.
What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!
[tex]\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2[/tex]
[tex]h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}[/tex]
[tex]h=-16t+64t+3[/tex]
We want to find out when it returns to ground level (h = 0)
To find this out, we can plug in 0 and solve for t.
[tex]0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}[/tex]
[tex]= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}[/tex]
So the ball will return to the ground at the positive value of [tex]\boxed{\frac{8+\sqrt{67}}{4}}[/tex] seconds.
What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!
[tex]\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2[/tex]
[tex]h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}[/tex]
