Question

A tennis ball rolls off the edge of a tabletop 0.550 m above the floor and strikes the floor at a point 1.30 m horizontally from the edge of the table.

Part A: Find the time of flight of the ball. ANSWER: 0.335 s

Part B: Find the magnitude of the initial velocity of the ball. ANSWER: 3.88 m/s

Part C: Find the magnitude of the velocity of the ball just before it strikes the floor. ANSWER: 5.08 m/s

Part D: Find the direction of the velocity of the ball just before it strikes the floor. ANSWER: ????????

Answer 1

You must begin by finding the time it takes to free-fall to the floor. This is independent of any horizontal velocity the object might have.
s = 0.550m = ½at² = ½•9.8m/s²•t²
t = 0.335s ← (A)

The initial velocity of the ball was entirely horizontal, and which can now be determined because we know how far it landed from the table and how long it took to do so.
Vh = s / t = 1.3m / 0.3348s = 3.88 m/s ← (B)

The easiest way to find the vertical velocity just before impact is through
Vv = at = 9.8m/s²•0.3348s = 3.281 m/s
So the resultant velocity = sqrt(3.281^2 + 3.88^2) m/s = 5.08 m/s ← (C)

Θ = arctan(Vv / Vh) = arctan(3.281/3.88) = 40.21º ←(D)