Velocity just before hitting the sand=under root of 2gh =under root of 2X9.8X1 =1.4m/s

Let F be the resistance force,so net force on the body after it start penetrating into the sand is F-mg upward(de acceleration will take place)

deacc of the body=(F-mg)/m =(F/m)-g

square of v -square of u = 2as --------- equation no-1 now v=0 as boby finally comes to rest u(initial velocity)=1.4m/s a=(f/m)-g s=1m

put these values in equation 1

Here F is unknown After solving the equation F (resistance Force)cmoes out to be 10780 Newton

BASHEER AHAMED. H

Aug 6th, 2012

No, the above answer is partially correct.
Look,
Just before reaching the ground, the velocity of the body is as calculated above (1.4 m/s). Once it start penetrating it into the ground, it will have new acceleration a and not the acceleration due to gravity.

To calculate it, we have Newtons equations of motion. ie, v^2 = u^2 + 2as.

Given that penetration depth = 1m.
Final velocity, v=0 m/s
Initial velocity (while penetrating sand) = 1.4 m/s
Finally, we get, a = -9.81 m/s^2.

First of all we have to calculate the potential energy of mass(100kg) just when it is going to fall. As we know that energy cannot be created nor destroyed, so, only one forms of energy changes to other. Here, potential energy of mass (when it is 1 m above the ground---actually we can take it 2 m because mass comes in rest after traveling 1m below the sand surface) is converted to kinetic energy and which finally converts into heat energy by the resistance force created by the sand. So, potential energy (initially) = mgh=100x10x2=2000j. Since, the same energy is completely exhausted during deceleration (in sand), (we are here assuming that sand exerts a constant resistance force (F),so, we can calculate value of this force as follows...

Work done by this resistive force = F.S=Initial potential energy=2000 J
here, S=displacement of mass when resistance force is acting (or total travel in sand)=1m

so, F=2000/1 =2000 N (ans.)

It seems to be impossible for any body falling from a height of 1 m will go 1m inside the sand against the heavy resistance offered by the sand.
So, my answer may be correct theoretically but it is not practically true (too low value--2000 N of resistance offered by sand)

KINGSHUK

Feb 15th, 2015

sqrt(2*g*h)= 4.4 m/s not 1.4

Tarek

Mar 3rd, 2015

http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html
Please check this page.
it shows a reasonable answer.

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