Question

How do I find the Velocity and How long will it take for the ball to reach it's maximum height?

Answer 1

So hmm check the picture below

$\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\ \\ \begin{cases} v_o=\textit{initial velocity of the object}\to &64\\ h_o=\textit{initial height of the object}\to &12\\ h=\textit{height of the object at "t" seconds} \end{cases}\\\\ -----------------------------$

$\bf \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} h(t)=&-16t^2&+64t&+12\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

part 1)  

it takes  $\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds$

part 2)

$\bf \textit{now, doubling }v_o\\\\ \begin{cases} v_o=\textit{initial velocity of the object}\to &128\\ h_o=\textit{initial height of the object}\to &12\\ h=\textit{height of the object at "t" seconds}\end{cases}\\\\ -----------------------------\\\\ \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} h(t)=&-16t^2&+128t&+12\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

it will reach the maximum height at   $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet$


how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$ ?

subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became