Question

a ball is thrown straight up from a rooftop 192 feet high. the formula below describes the balls height above the ground, h, in feet, t seconds after it was thrown. the ball misses the rooftop on its way down. the graph of the formula is shown. determine when the balls height will be 96 feet and identify the solution as a point on the graph.
h = - 16t² + 40t + 192
the balls height will be 96 feet at second(s).

Explanation:

Step1: Set up the equation

Set $h = 96$ in the equation $h=-16t^{2}+40t + 192$. So we get $96=-16t^{2}+40t + 192$.

Step2: Rearrange to standard quadratic - form

Move all terms to one side: $16t^{2}-40t - 96=0$. Divide through by 8 to simplify: $2t^{2}-5t - 12 = 0$.

Step3: Factor the quadratic equation

Factor $2t^{2}-5t - 12$ as $(2t + 3)(t - 4)=0$.

Step4: Solve for t

Set each factor equal to zero:
If $2t+3 = 0$, then $2t=-3$, and $t=-\frac{3}{2}$.
If $t - 4=0$, then $t = 4$. Since time $t\geq0$ in this context, we discard $t=-\frac{3}{2}$.

Answer:

$4$