Question

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long does the dive last and with what speed does the person enter the water

Answer 1

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

    The height of the cliff is  $h = 33 \ m$

 Generally from kinematic equation we have that

       $h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is  u =  0 m/s

 So

        $33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=>      $t = 2.595 \ s$

Generally from kinematic equation

     $v= u + gt$

=>  $v= 0 + 9.8 * 2.595$

=>  $v = 25.43 \ m/s$