A rock is thrown horizontally off a cliff with an initial speed of 3 m/s. The initial position of the rock is 10 meters above the ground. How long is the rock in the air?

1.43 s.

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