Question

21 POINTS!!!!! The height of a projectile can be described as follows
y = −16t^2 + v0t + h0

Where t=time in seconds;
y= height in feet
v0=initial velocity
h0=initial height
What would be the maximum height for a squirrel that jumped from a branch that was 10 ft off the ground with an initial velocity of 12 ft/s.

Answer 1

check the picture below

$\bf \text{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\ \\ \quad \\ \begin{cases} v_o=\textit{initial velocity of the object}\to &12\\ h_o=\textit{initial height of the object}\to &10\\ h=\textit{height of the object at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+12t+10\\\\ -----------------------------$

$\bf \textit{where's the vertex of it? well} \\\\\\ \textit{vertex of a parabola}\\ \quad \\ y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$


so... that's the squirrel's maximum height, at $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$   feet