Question

I need help with this awnser i’m really lost

Answer 1

check the picture below.


`\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{48}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at`

so the ball will reach its highest at the "vertex", let's check then.


`\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+48}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{48}{2(-16)}~~,~~6-\cfrac{48^2}{4(-16)} \right)\implies \left(\cfrac{3}{2}~~,~~6+36 \right) \\\\\\ \left( \stackrel{\textit{how many seconds}}{1(1)/(2)}~~,~~ \stackrel{\textit{maximum height}}{42} \right)`