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Newtons laws- object sinking into ground -help needed

An object with a mass of 200 grams is thrown upwards from the top of a 100m building with an initial speed of 21 ms.

The ground is soft and the object sinks into it. If the ground exerts a constant resistive force of 2500N, how far does the object sink into the ground?

I have worked out the max height is 22.5m therefore the total height from max to the ground is 122.5m also the speed it hits the ground I have calculated to be 49ms.

I don't know how to work out how far into the ground it will go.

Any help greatly appreciated!! Thannnnnks!!
Reply 1
Original post by Mechanicsstudent
An object with a mass of 200 grams is thrown upwards from the top of a 100m building with an initial speed of 21 ms.

The ground is soft and the object sinks into it. If the ground exerts a constant resistive force of 2500N, how far does the object sink into the ground?

I have worked out the max height is 22.5m therefore the total height from max to the ground is 122.5m also the speed it hits the ground I have calculated to be 49ms.

I don't know how to work out how far into the ground it will go.

Any help greatly appreciated!! Thannnnnks!!

You know the KE when it hits the ground so can equate that to the work done against friction when it sinks into the ground. I guess you should also include the change in GPE, but I doubt its substantial.
(edited 6 months ago)
I think I've worked it out!! 0.096m
Reply 3
Looks about right. To get the ground speed you could have done it directly using v^2 = u^2+2as, but it wouldnt be that much simpler. In a sense it would be equivalent to saying that youre summing the iniital GPE and initial KE to get the KE when it hits the ground.

Original post by Mechanicsstudent
I think I've worked it out!! 0.096m
(edited 6 months ago)
Original post by Mechanicsstudent
I have worked out the max height is 22.5m therefore the total height from max to the ground is 122.5m also the speed it hits the ground I have calculated to be 49ms.


Unless you were asked for the max. height, it's a waste of time/energy to work it out, and it introduces more potential for error with the extra calculation.

suvat applies as long as the acceleration is constant throughout the motion. From being thrown to hitting the ground that is the case.

So, v2=212+2×9.8×100v^2= 21^2+ 2\times 9.8\times 100, and v=49v=49 m/s
(edited 6 months ago)

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