physics

A 65 kg diver jumps off of a 10.0 m tower.

a. Find the diver's velocity when he hits the water.

b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.

Ok, im confused, I found velocity which is 14. But for the net force im confused on what we are finding when using this equation below.

For the force, it would be similar to the car acceleration problem, except that to find the deceleration in water, we would use
v1²-v²=2aS
S=distance (2 m)
v1=final velocity =0
v=initial velocity as found above
a=deceleration due to water (negative) including effects of buoyancy and gravity.
and finally
F=ma

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  1. You're quite right, the deceleration as calculated above accounts for gravity, but the force does not.
    From v=14, v1=0, we calculate a=(14²-0²)/2/2m=49
    F=ma=65*49=3185 N
    On this we have to add the weight of the diver, 65*9.8 to give 3822 N.

    As suggested by Drwls, it is simpler to do this all in one step by energy considerations, if that's the topic you are working on in class.

    Σ PE+KE at start
    = Σ PE+KE+work done
    PE=potential energy
    KE=kinetic energy

    At start:
    PE = mgh (h=12 m relative to final position)
    KE = 0
    At end:
    PE = 0
    KE = 0
    Work done = F*2 metres
    Therefore
    mgh+0 = 0 + 0 + 2F
    F=mgh/2
    = 65*9.8*12/2
    = 3822 N

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  2. bob

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