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Does the amount of force an object experience at a high height in a waterfall is same as the amount of force the object experience at a lower height in waterfall? Does change in kinetic energy also lead to change in amount of force applied by object?

Why does it create a much damage when a heavy stone dropped from a height on the ground than dropped from a low height. I know momentum change! But the amount of force applied by object on ground should be same no?, why height matters!i.e F=ma (m-mass of stone a-9.8m/s^2)?

I think impact should depend on Amount of force applied by object? Or momentum? I can't figure it out, please clear my misconception!

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I think impact should depend on Amount of force applied by object? Or momentum?

Yes, if by impact you mean force, then the force experienced by the object that is hit depends on the momentum of the colliding object.

There is a formula from Newton's 2nd law, Force x time = change in momentum. If the colliding object is brought to rest $$Ft=mv$$ or $$F = \frac{mv}{t}$$

For the waterfall or falling water example, let's consider the following.

Pouring water from a jug at constant rate (kilograms per second) and stopping it with your other hand at different heights.

The $\frac{m}{t}$ part of the equation is the same, but the velocity of a falling object increases as it falls. Your hand experiences more force from the water when it's further underneath the jug simply due to the $v$ being greater further down.

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  • $\begingroup$ I cannt get it!! See my query in @java monke comment $\endgroup$
    – user315475
    Oct 16, 2021 at 15:59
  • $\begingroup$ And also does the amount of force, ball apply in both case on the ground is same? (When hitted) i have one more query may be i am misunderstanding that when we say amount of force applied by object on earth or ground is F=ma. What does it mean? Does it mean that object is attracting(force direction is perpendicularly upward to earth) earth with same force as earth attracting object OR does it mean F=ma is the amount of force supplied by object (perpendicular downward direction of force to earth) when hit the ground?hope u can get it $\endgroup$
    – user315475
    Oct 16, 2021 at 16:23
  • $\begingroup$ Please check out 3 of my queries sir! Please $\endgroup$
    – user315475
    Oct 16, 2021 at 16:23
  • $\begingroup$ @ Anshika Singh In F=ma, a is the deceleration. The deceleration is change in speed / change in time. If the time is the same to stop the object e.g. 0.1 second, then if the object has falled further and reached a higher speed e.g. 20m/s instead of 10m/s, then the deceleration is 200m/s^2 instead of 100m/s^2 and for a 1kg mass F=ma gives 200N instead of 100N. i.e the floor must exert more force to stop the object and from Newton's 3rd law the object exerts more force on the floor $\endgroup$
    – John Hunter
    Oct 16, 2021 at 19:24
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    $\begingroup$ @ Anshika Singh Yes, you seem to be ok with it now, all the best $\endgroup$
    – John Hunter
    Oct 17, 2021 at 8:20
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Suppose the ball is at a certain height and the ball is dropped, when it takes time t to reach the ground and its final velocity just before hitting the ground is v. Gravitational acceleration here is g.

Using the equation of motion v=u+at, we get v = gt, hence we can say that the velocity increases with time.

If the ball was at a greater height, it would have taken more time for that ball to reach the ground.

Now according to Newton's Second Law of Motion,

F ∝ Rate of Change of Momentum →(i)

Let Pᵤ be initial moment and Pᵥ be final momentum, and t be time, then

​F ∝ (Pᵥ-Pᵤ)/t

or F = k{(Pᵥ-Pᵤ)/t}

Let final velocity be vᵥ and initial velocity be vᵤ.

F = k{(Pᵥ-Pᵤ)/t}

= k{(mvᵥ-mvᵤ)/t}

= k[{m(vᵥ-vᵤ)}/t]

And since acceleration(a) is rate of change of velocity, a=(vᵥ-vᵤ)/t, hence, by substituiton,

F = k(ma) →(ii) and in the SI system of units, k=1; hence F=ma.

From equations (i) and (ii), we can see than F∝a and F ∝ Final Momentum.

I think impact should depend on Amount of force applied by object? Or momentum?

Greater the force, greater the final momentum of the body, hence, the impact depends on both force and momentum as they are mathematically related.

P.S.: MathJax wasn't working, hence the equations are not exactly aesthetically pleasing, any kind edits would be highly appreciated.

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  • $\begingroup$ Yah, if we see like this, everything seem fine but.... If we look this this,that a ball say 1kg and drop from a 2m from ground. Than Force it apply on ground will be F =ma i e 1*9.8= 9.8N. Similarly if that ball is held 5m from ground than also force it apply on ground F=ma, 1*9.8=9.8N again! So if look this impact should be same no? But look through monentum perspective final momentum in 1 case will be p=mv, p= 1* 6.26 (u=0, a=9.8 s=2m calculate by v^2-u^2=2as). Similarly if calculate in case 2 p =1*9.8 (u=0,a=9.8,s=5m calculated from v^2-u^2=2as). See! P1=6.26, P2=9.89 it tells.... $\endgroup$
    – user315475
    Oct 16, 2021 at 15:27
  • $\begingroup$ It tells case 2 will have more impact!..... Isnt that weird!! Or i am wrong some where. Also... If we look from Time percpective case 1 F=(mv1-mv2)/t (t will be less but mv2 will also be less n!! ), case 2 F=(mv1-mv2)/t (here t will be more than t in case 1 as ball is at height but along with it mv2 will also be more!! As velocity will increase with time) , so overall we can say mu1-mv2/t in case 1 will be equal mu1-mv2/t in case 2. Thus a1=a2=9.8m/s2 approx. So f1=f2= (m*a) so impact should be same. According to this momentum view show something else while force view show sonething else?!! $\endgroup$
    – user315475
    Oct 16, 2021 at 15:39
  • $\begingroup$ If you are asking about the "impact", it basically means force, and for the cases(heights) presented above, the ratio of the forces or IMPACTS of the ball on the ground will depend on the Force-Impulse Equation of Newtons 2nd Law, which has been shown in the other answer by @John Hunt. Check that answer out. $\endgroup$
    – AltercatingCurrent
    Oct 16, 2021 at 15:42

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