Answer:
v_o= 12 m/s
Explanation:
The ball experienced a constant acceleration motion, so we need to apply the following formula:
[tex]y=y_o+v_o*t+\frac{1}{2}*a*t^2\\\\vo=\frac{y-y_o-\frac{1}{2}*a*t^2}{t}\\\\v_o=\frac{0m-8m-\frac{1}{2}*(-9.81m/s^2)*(3.00s)^2}{3.00s}\\v_o=12m/s[/tex]
Note: we set the acceleration of gravity as negative because it is going down.
the ball was thrown vertically upward with an initial velocity of 12 m/s
