Question

Is anyone good at math? I need help solving this problem.

A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be?

Answer 1

so hmm check the picture below

$\bf \text{initial velocity}\\ h(t) = -16t^2+v_ot+h_o \qquad \text{in feet}\\ \\ \quad \\ \begin{cases} v_o=\textit{initial velocity of the object}\to &24\\ h_o=\textit{initial height of the object}\to &50\\ h=\textit{height of the object at "t" seconds} \end{cases}\\\\ -----------------------------\\\\ \textit{vertex of a parabola}\\ \quad \\ y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so.. the ball will reach its maximum height at $\bf -\cfrac{{{ b}}}{2{{ a}}}$ seconds

and will be $\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}$ feet above the ground