physics

An 70 kg man drops from rest on a diving
board −2.9 m above the surface of the water
and comes to rest 0.58 s after reaching the
water.
The acceleration due to gravity is
9.81 m/s
2
.
What force does the water exert on the
man?
Answer in units of N

  1. 👍 0
  2. 👎 0
  3. 👁 336
asked by allioe
  1. I think it would just be Fg x m, so 70kg times 9.81 m/s^2, and then multiply all of that by 2.9m, and the negative is just saying what direction the man is falling, meaning downwards

    1. 👍 0
    2. 👎 0
    posted by K
  2. mgh=mv²/2
    v=sqrt{2gh} =sqrt{2•9.8•2.9} =7.5 m/s
    v(fin)=v-at
    v(fin)=0
    a=v/t
    x=vt -at²/2=vt/2
    mv²/2 =W(fr) = F(fr) •x =>
    F(fr)= mv²/2x = 2mv²/2 vt=mv/t =
    =70•7.5/0.58 = 905.2 N

    1. 👍 0
    2. 👎 0
    posted by Elena
  3. The above equation is wrong

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    2. 👎 0
    posted by T

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