Final answer:
The velocity of the ball just before it hits the ground is approximately 34.3 m/s.
Step-by-step explanation:
To find the velocity of the ball just before it hits the ground, we need to consider the vertical component of the ball's motion. Since the ball is thrown horizontally, the initial vertical velocity is zero. We can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (which is -9.8 m/s^2 due to gravity), and s is the vertical distance travelled, which in this case is the height of the building (60 m). Rearranging the equation, we get v^2 = 0 + 2(-9.8)(60), which simplifies to v^2 = -1176. Solving for v, we find that the velocity of the ball just before it hits the ground is approximately 34.3 m/s.