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V125BC [204]
3 years ago
13

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

oes the dive last and with what speed does the person enter the water
Physics
1 answer:
USPshnik [31]3 years ago
3 0

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

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