A ball falls 0.35m to the floor. it lands 3m from the edge of the table. what was the velocity of the ball before it left the table?

The ball rolls off the table with speed v from a height of 0.35 m, so that it covers a horizontal distance x with height y at time t of

x = v t

y = 0.35 m - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 0, i.e. the time it takes for the ball to reach the ground:

0 = 0.35 m - 1/2 g t ²

t ² = (0.70 m) / g

t ≈ 0.267 s

Now solve for v given that the ball falls 3 m away from the table:

3 m = v (0.27 s)

v = (3 m) / (0.27 s)

v ≈ 11.2 m/s

the answer would be c