Answer:
1.43 s.
Step-by-step explanation:
Using one of the equations of motion,
S = ut + 1/2gt².......................... Equation 1
Where S = height of the cliff, u = initial velocity, t = time, g = acceleration due to gravity.
Note: When the rock begins to fall from the maximum height, u = 0 m/s, g = positive
Given: S = 10 m, u = 0 m/s
Constant: g = 9.8 m/s²
Substituting these values into equation,
10 = 0(t) + 1/2(9.8)(t²)
10 = 0 + 4.9t²
t² = 10/4.9
t² = 100/49
t = √(100/49)
t = 10/7
t = 1.43 s.
Thus the rock spend 1.43 s in air