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If a ball is falling under free fall then the force exerted by the ball on the ground would be $mg$. But that's not the case in real life ball would hit with more force. But when i draw free body diagram there is only one force that is acting on it $mg$

Can someone explain what is the force the ball will exert ?

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  • $\begingroup$ How does it have more force than it's mass times the acceleration of gravity? $\endgroup$ – Adrian Howard Oct 18 at 6:37
  • $\begingroup$ you forgot the force that is needed to slow down the ball. $\endgroup$ – Umaxo Oct 18 at 7:34
  • $\begingroup$ I wrote free fall $\endgroup$ – Harshit Gupta Oct 18 at 7:50
  • $\begingroup$ This means that there is negligible air resistance $\endgroup$ – Harshit Gupta Oct 18 at 7:50
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Interesting question. Ball force affecting ground is $$ F_c = \frac{\Delta p}{t_i} = \frac{mv}{t_i} $$, because end speed is zero.

Now let's assume ball started to fall from some altitude, so it reached it's "contacting" speed $v$ according to : $$ v = v_0 + a \space t = g \space t_f $$

Plunging this into formula above, we get :

$$ F_c = \frac{mg \space t_f}{t_i} = W \frac{t_f}{t_i} $$

So ball force affecting ground is it's weight scaled by falling time over interaction time with ground ratio.
Analogically ground will exert normal force with same magnitude on ball. That's why eggs breaks falling down :-)

Only in case when $ t_f \approx t_i $, i.e. when falling time approaches interaction time, contact force equals weight. This principle is exploited in various kinds of trampolines and Life net used by firefighters.

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  • $\begingroup$ So if $t_i$ goes to zero then the force would be infinte ? $\endgroup$ – Harshit Gupta Oct 18 at 7:52
  • $\begingroup$ Mathematically - yes. But physically, smallest meaningful time duration is plank time $t_p$. Interaction time can't be infinitesimally small, because all interactions can go only at maximum speed of light $c$. $\endgroup$ – Agnius Vasiliauskas Oct 18 at 8:01
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The ground must do work on the falling ball to bring it to a stop. From the work energy theorem the net work done on the ball equals its change in kinetic energy or

$$F_{Ave}d=\frac{mv^2}{2}=mgh$$

$$F_{Ave}=\frac{mgh}{d}$$

Where $F_{Ave}$ = average impact force, $d$ = stopping distance after impacting the ground due to deformation of the ground and/or ball, and $h$ = height from which the ball dropped. Assumes $d$ << $h$ to neglect the change in potential energy after impact

Hope this helps

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the force exerted by the ball on the ground would be $mg$

Why do you say that?

The force that the ball exerts on the ground depends on the modulus of elasticity (stiffness) of the ground and can be very high. The normal reaction of the ground stops the ball over a certain distance. If the modulus of elasticity of the ground is high, the stopping distance is short and the force of reaction is high.

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perhaps you also want to calculate the contact force between the ball and the ground.

your ball is falling from height $H$ with start velocity $v_0$ and then touch the ground. if we take a simple model of the ground with stiffness $k$ and damper $d$ you get this equation of motion.

$$m\ddot{x}+d\,\dot{x}+k\,x=m\,g\tag 1$$

your contact force is the spring plus the damper force (free body diagram)

$$F_c=d\,\dot{x}+k\,x$$

to find the solution of equation (1), you need the initial conditions:

$x(t=0)=0$

$\dot{x}(t=0)=v_e$

where $v_e$ is the velocity of your ball when it touch the ground:

$v_e=\sqrt{v_0^2+2\,g\,H}$

where $v_0$ is the start velocity of ball.

This diagram shows the contact force :

enter image description here

the blue line is the weight of the ball. the red one is the start velocity $v_0=0$ and the green one the start velocity is $v_0=10$

you see that your contact force is much higher then the weight force and is depending on the velocity $v_e$ which is a function of the height $H$ and the start velocity $v_0$

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