Force changes as object falls?

[merced]

force changes as object falls??

"A high diver of mass 70.0 kg jumps off a board 10.0 m above the water. If his downward motion is stopped 2.00 s after he enters the water, what average upward force did the water exert on him?"

Do I need to find the velocity when the person hits the water?
Does force change as velocity changes, i.e. does the man hit the water at a greater force if he falls from a greater height? (It seems force should increase as the height increases, but force = mass * acceleration, and has nothing to do with velocity, right?)

 

[semc]

i believe u need to consider that force is rate of change of momentum with respect to time and since momentum is mv,velocity does play a part right?

 

[Dorothy Weglend]

"A high diver of mass 70.0 kg jumps off a board 10.0 m above the water. If his downward motion is stopped 2.00 s after he enters the water, what average upward force did the water exert on him?"

Do I need to find the velocity when the person hits the water?
Does force change as velocity changes, i.e. does the man hit the water at a greater force if he falls from a greater height? (It seems force should increase as the height increases, but force = mass * acceleration, and has nothing to do with velocity, right?)

If you hit the water at a higher velocity, then probably the acceleration caused by the water will be greater (that all depends on the time it takes to get to zero v of course). And as you said, F = ma.

Dorothy

 

[SGT]

"A high diver of mass 70.0 kg jumps off a board 10.0 m above the water. If his downward motion is stopped 2.00 s after he enters the water, what average upward force did the water exert on him?"

Do I need to find the velocity when the person hits the water?
Does force change as velocity changes, i.e. does the man hit the water at a greater force if he falls from a greater height? (It seems force should increase as the height increases, but force = mass * acceleration, and has nothing to do with velocity, right?)

velocity = acceleration * time.
If the diver jumps from a higher point he will take more time to accelerate and his final velocity will be greater.
To calculate the average upward force needed to stop him in 2 s, you need to know his velocity when he hits the water.

 

[semc]

may i ask why u can use velocity = acceleration * time in this case since the qn didnt state that air resistance is neligible?

 

[Astronuc]

One may solve the problem by neglecting air resistance, which is negligble at low velocities, and one should probably state that as part of the solution. If this is an introductory course in physics, that is likely the case.

One certainly does not hit 'terminal velocity' dropping from 10 m.

v = a * t, as long as a is 'contant'. If a = a(t) then one must integrate appropriately.

See definition of acceleration - http://hyperphysics.phy-astr.gsu.edu/hbase/acca.html

 

[SGT]

As Astronuc explained, air resistance is negligible at low velocities.If we considered air drag the problem would be very complex. The drag force is given by
$$D = \rho S C_D v^2$$
Where $$\rho$$ is air density.
S is the cross section of the body perpendicular to the velocity. In the case of a diver, this cross section varies with time, since the diver starts with his/her head up and finishes with it pointing down.
$$C_D$$ is the drag coefficient for the body. I have no idea where to find it for a human body.
v is the velocity.
If you could obtain the drag coefficient, you would have to model the rotation of the body of the diver in order to solve the problem iteratively. Hardly the kind of problem for an introductory physics course.