Displacement-Time Graph of a bouncing ball

Then when you square the time the graph becomes linear because the displacement is now linearly dependent on time.initially you talked of...initial velocity being zero. so in order to have a linear graph you need to have a zero initial velocity.
  • #1
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Homework Statement


Does anyone know how to plot a displacement-time graph that will give the average velocity of a bouncing ball? The ball bounces 5 times and I would like to know it's average velocity. What equations would I use?

Homework Equations


0.5mv^2=mgh?

The Attempt at a Solution


I simply plotted displacement vs time graph but I get a quadratic (or exponential) function. I just want to confirm if I am doing this right. Thank you.
 
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  • #2
Jeven said:
Does anyone know how to plot a displacement-time graph that will give the average velocity of a bouncing ball? The ball bounces 5 times and I would like to know it's average velocity. What equations would I use?

the bouncing ball is a repetitive event and in between the bounces the speed increase -goes to a maximum and then reverses its path. so the displacement time graph should be quadratic in nature.
what is meant by average velocity -i fail to understand but if you take the slope of displacement time graph at various points of time , then a graph of velocity and time can be plotted and one can try to get average velocity.
 
  • #3
drvrm said:
the bouncing ball is a repetitive event and in between the bounces the speed increase -goes to a maximum and then reverses its path. so the displacement time graph should be quadratic in nature.
what is meant by average velocity -i fail to understand but if you take the slope of displacement time graph at various points of time , then a graph of velocity and time can be plotted and one can try to get average velocity.

So what can I do to my graph to make it a linear function?
 
  • #4
Jeven said:
So what can I do to my graph to make it a linear function?

a displacement can depend linearly on time only in case of motion with uniform velocity- so the physical event of bouncing ball is not a system having uniform velocity.
therefore how one can force the system to change its action.

however if you plot displacement with square if time - artificially one can show a linear dependence- and for that to happen the initial velocity must be zero.
 
  • #5
drvrm said:
a displacement can depend linearly on time only in case of motion with uniform velocity- so the physical event of bouncing ball is not asystem of uniform velocity.
therefore how one can force the system to change its action.
however if you plot displacement with square if time - artificially one can show a linear dependence- for that the initial velocity must be zero.
Can you explain why squaring the time would result in a graph where the gradient is the velocity? Also, would displacement be displacement^2 or stay displacement?
 
  • #6
Jeven said:
Can you explain why squaring the time would result in a graph where the gradient is the velocity? Also, would displacement be displacement^2 or stay displacement?

well one should look at the physical event -
say a ball falling on floor from a
height say h initial velocity =0
so write down equation for displacement - it should be a function of t^2;
again after hitting the floor and bouncing back with finite velocity the displacement will be a function of time and time square.
 
  • #7
drvrm said:
well one should look at the physical event -
say a ball falling on floor from a
height say h initial velocity =0
so write down equation for displacement - it should be a function of t^2;
again after hitting the floor and bouncing back with finite velocity the displacement will be a function of time and time square.
Ok so basically, if I graph time squared and displacement, I WILL get the velocity from the gradient is what you are saying?
 
  • #8
Jeven said:
Ok so basically, if I graph time squared and displacement, I WILL get the velocity from the gradient is what you are saying?

yes, that's my hint.
 
  • #9
drvrm said:
yes, that's my hint.
But then I have an equation t^2=2s/a ... so where is the velocity?
 
  • #10
Jeven said:
But then I have an equation t^2=2s/a ... so where is the velocity?
for velocity you have the displacement -time graph and if you take slope i.e. rate of change of displacement with time then only you can get velocity -
from any other graph you do not get it as velocity is unique.
 
  • #11
drvrm said:
for velocity you have the displacement -time graph and if you take slope i.e. rate of change of displacement with time then only you can get velocity -
from any other graph you do not get it as velocity is unique.

Then what happened to the time squared? I want to find the velocity of a bouncing ball as it bounces 5 times. What do I do?
 
  • #12
Jeven said:
I want to find the velocity of a bouncing ball as it bounces 5 times. What do I do?

initially you talked of average velocity-
now you say the velocity which can bounce five times.
so clarify your question again.
 
  • #13
drvrm said:
initially you talked of average velocity-
now you say the velocity which can bounce five times.
so clarify your question again.

My question is what is the average velocity of a ball after 5 bounces?
 
  • #14
Jeven said:
My question is what is the average velocity of a ball after 5 bounces?

then you have to work with displacement -time quadratic curve.
if however you have an ideal ball and ideal collision -elastic the one can predict the velocity after any number of bounces as the linetic energy is not lost in bouncing-
i.e. if it is dropped from a height h
the final velocity before the hit will be mgh =1/2 . m v^2 ; so v= sqrt(2.g.h)
and after elastic collision its speed will be same in opposite direction.
so after 5th bounce also it should have the same velocity.
 
  • #15
drvrm said:
then you have to work with displacement -time quadratic curve.
if however you have an ideal ball and ideal collision -elastic the one can predict the velocity after any number of bounces as the linetic energy is not lost in bouncing-
i.e. if it is dropped from a height h
the final velocity before the hit will be mgh =1/2 . m v^2 ; so v= sqrt(2.g.h)
and after elastic collision its speed will be same in opposite direction.
so after 5th bounce also it should have the same velocity.

Yeah but my ball is not elastic, it jumps to around 0.5 of its initial height. What do I do to this quadratic curve to get a straight line whose gradient is average velocity?
 

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