Question

if a ball is thrown directly upward with a velocity of 40 ft/s, its height (in feet) after t seconds is represented by 40t-16t^2. what is the maximum height attained by the ball?

Answer 1

check the picture below.


`\bf h(t)=40t-16t^2\implies h(t)=-16t^2+40t+0 \\\\\\ \begin{array}{llll} \qquad \textit{in feet}\\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\textit{initial velocity of the object}\\ \qquad 40\\ h_o=\textit{initial height of the object}\\ \qquad 0\\ h=\textit{height of the object at`


`\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\ \begin{array}{llll} y = &{{ -16}}x^2&{{ +40}}x&{{ +0}}\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right) \\\\\\ \left( \stackrel{\textit{it took this long}}{-\cfrac{40}{2(-16)}}~~,~~\stackrel{\textit{to get this high}}{0-\cfrac{40^2}{4(-16)}} \right)\qquad \stackrel{\textit{maximum height}}{0-\cfrac{40^2}{4(-16)}}`