consider the motion of the ball in vertical direction after it rolls off the table
v₀ = initial velocity of the ball in vertical direction = 0 m/s
a = acceleration = acceleration due to gravity = 9.8 m/s²
Y = vertical displacement = height of the table = 0.91 m
t = time spent in air by the ball
Using the kinematics equation
Y = v₀ t + (0.5) a t²
0.91 = 0 t + (0.5) (9.8) t²
t = 0.43 sec
v = final velocity in vertical direction just before it hits the ground
Using the kinematics equation
v = v₀ + at
v = 0 + (9.8) (0.43)
v = 4.21 m/s
