Explanation:
Using the equation of motion S = ut+1/2gt² where;
S is the height of the table = 0.60m
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.81m/s²
t is the time taken
Substitute
0.6 = 0+1/2(9.81)t²
0.6 = 4.905t²
t² = 0.6/4.905
t² = 0.1223
t = √0.1223
t = 0.3497
t≈0.35secs
Hence it will take 0.35seconds for a ball launched off the edge of these tables to reach the floor
