Answer:
The time is [tex]t = 2.595 \ s[/tex]
The speed is [tex]v = 25.43 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the cliff is [tex]h = 33 \ m[/tex]
Generally from kinematic equation we have that
[tex]h = ut + \frac{1}{2} gt^2[/tex]
before the jump the persons initial velocity is u = 0 m/s
So
[tex]33 = 0 * t + \frac{1}{2} 9.8 * t^2[/tex]
=> [tex]t = 2.595 \ s[/tex]
Generally from kinematic equation
[tex]v= u + gt[/tex]
=> [tex]v= 0 + 9.8 * 2.595[/tex]
=> [tex]v = 25.43 \ m/s[/tex]
