Answer:
<em>1.43 s.</em>
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.
<em>Note: When the rock begins to fall from the maximum height, u = 0 m/s, g = positive</em>
<em>Given: S = 10 m, u = 0 m/s</em>
<em>Constant: g = 9.8 m/s²</em>
<em>Substituting these values into equation,</em>
<em>10 = 0(t) + 1/2(9.8)(t²)</em>
<em>10 = 0 + 4.9t²</em>
<em>t² = 10/4.9</em>
<em>t² = 100/49</em>
<em>t = √(100/49)</em>
<em>t = 10/7</em>
<em>t = 1.43 s.</em>
<em>Thus the rock spend 1.43 s in air</em>
