Question

28. Someone throws a rubber ball vertically upward from the roof of a building 8.00 m in height. The ball rises,

Answer 1

Answer:

v_o= 12 m/s

Explanation:

The ball experienced a constant acceleration motion, so we need to apply the following formula:

$y=y_o+v_o*t+\frac{1}{2}*a*t^2\\\\vo=\frac{y-y_o-\frac{1}{2}*a*t^2}{t}\\\\v_o=\frac{0m-8m-\frac{1}{2}*(-9.81m/s^2)*(3.00s)^2}{3.00s}\\v_o=12m/s$

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