Question

a person kicks a ball from the ground into the air with an initial upward velocity of 8 feet per second. what is the maximum height of the ball? when will the ball hit the ground?

Answer 1

Chech the picture below.

if the person kicked it from the ground, that means its initial height is 0.

it reaches its maximum height at the y-coordinate of its vertex, and it will hit the ground when y = 0, as you see in the picture.

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{8}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+8t+0$

now let's find the y-coordinate of its vertex

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+8}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(\qquad ,~~0-\cfrac{8^2}{4(-16)} \right)\implies (\quad ,~0+1)\implies (\quad ,~1)$

when will it hit the ground?

$\bf \stackrel{h(t)}{0}=-16t^2+8t\implies 16t^2-8t=0\implies 8t(2t-1)=0 \\\\\\ 8t=0\implies \stackrel{\textit{0 seconds when it took off first}}{t=0} \\\\\\ 2t-1=0\implies 2t=1\implies \stackrel{\textit{half a second later it when it came back down}}{t=\cfrac{1}{2}}$