Question

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?
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Answer 1

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