Question

I need help with a. b. and c.​

Answer 1

Check the picture below.

so if we just find the vertex of that parabolic path, we'd get how far up it went and how long it took

`\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+16}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)`

`\left(-\cfrac{ 16}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (16)^2}{4(-16)}\right) \implies \left( - \cfrac{ 16 }{ -32 }~~,~~0 - \cfrac{ 256 }{ -64 } \right) \\\\\\ \left( \cfrac{ -1 }{ -2 } ~~~~ ,~~~~ 0 +4 \right)\implies \stackrel{seconds ~~ feet }{\left(0.5~~,~~4 \right)}`

well, as you see in the picture, the horse hits the ground again when h(t) = 0

`\stackrel{h(t)}{0}=-16t^2+16t\implies 0=-16t(t-1)\implies t= \begin{cases} 0\\ 1 \end{cases}`

so at t=0 and t=1 it's touching the ground, at t=0 when it begins to jump at t=1 when it thumps back down, took that long.