Question

A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function h=-16t^2+64t+3
find the time it takes for the ball to return back to ground level.

find the maximum height rescued by the ball.

Answer 1

Here's our equation.

$h=-16t+64t+3$

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.

$0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}$

$= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}$

So the ball will return to the ground at the positive value of $\boxed{\frac{8+\sqrt{67}}{4}}$ 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!

$\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2$

$h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}$