Answer:
[tex]t=6.96s[/tex]
Explanation:
From this exercise, our knowable variables are hight and initial velocity
[tex]v_{oy}=96ft/s[/tex]
[tex]y_{o}=112ft[/tex]
To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft
[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]
[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]
Solving for t using quadratic formula
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]
[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]
Since time can't be negative the answer is t=6.96s
