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`


`\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`

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