If a steel ball were to fall, how far away would it land

[Thara]

Hello!
Imagine you drop a steel ball of weight x from a variable height y.

the ball hits a steel ramp on the ground which is at 45 degree's

I would like to know how I could work out how far away the ball will land.

I have had a good look around for this however the closest thing I can find is the height that the ball would bounce which is not what I am looking at :(

 

[BvU]

Hello Thara,

Here at PF, the guidelines require homework questions to be posted in the homework section. A very useful template there helps you get started:

Homework Statement

how far away from where?​

Homework Equations

What equations could help you out here ?​

The Attempt at a Solution

the closest thing I can find

Find ? You mean calculate ?​

 

[Thara]

Hello Thara,

Here at PF, the guidelines require homework questions to be posted in the homework section. A very useful template there helps you get started:

Homework Statement

how far away from where?​

Homework Equations

What equations could help you out here ?​

The Attempt at a Solution

Find ? You mean calculate ?​

This is not a homework question otherwise I would have posted it in the homework forum.

How far away from the place it was dropped.

I was considering the SUVAT equation but I didnt get very far with it.

 

[BvU]

This is not a homework question otherwise I would have posted it in the homework forum.

Ok, just curiosity then ?

To make a real exercise out of this we change

the ball hits a steel ramp on the ground which is at 45 degree's

to:
bounces elastically off a steel ramp on the ground which is at 45 degrees wrt horizontal. Disregard any change in rotation of the ball.

Still need the suvat equations, but have to add something for the collision. I propose something like: speed of the ball remains the same, direction changes just like with a ray of light on a mirror.

PS the thread is still moved to HW because that's where it belongs

 

[Thara]

I am investigation exclusion zones on building sites,

I have attached my attempt at working it out, The answer that I get is about 4 times greater that the preferable answer.. I am unsure what factors I could use to reduce this?

 

[jbriggs444]

I am investigation exclusion zones on building sites,

I have attached my attempt at working it out, The answer that I get is about 4 times greater that the preferable answer.. I am unsure what factors I could use to reduce this?

As I read your work, you have used theta ("$\theta$") as the angle at which the steel ball rebounds. You have assumed that theta is equal to 45 degrees. But per the givens of the problem, the steel plate on the ground is at 45 degrees. For an energy-preserving bounce, the angle of incidence (measured from the perpendicular) will be equal to the angle of reflection (again, measured from the perpendicular). With that in mind, can you calculate the angle at which the steel ball will bounce from a 45 degree plate?

What would the plate angle have to be to get a bounce at a 45 degree angle?

 

[Thara]

would that be 135 degrees?

 

[jbriggs444]

would that be 135 degrees?

If 45 degrees is at a slant facing East then 135 degrees would be at a slant facing West.

 

[Thara]

If 45 degrees is at a slant facing East then 135 degrees would be at a slant facing West.

I'm not getting where you are going with this

 

[jbriggs444]

I'm not getting where you are going with this

If a steel ball falling vertically hits a plate that is angled at 45 degrees from the horizontal, it strikes that plate at an angle that is 45 degrees away from being perpendicular, right?

At what angle will it rebound?

 

[Thara]

will it be perpendicular to the ground? I.e. no y direction?

 

[jbriggs444]

will it be perpendicular to the ground? I.e. no y direction?

Parallel to the ground, rather. Yes, it will have zero velocity in the y direction.

 

[BvU]

Dear thara,

Make a drawing !

Incoming velocity v_i has a component along the mirror (left blue) that is unchanged (lower green)
And a component along the perpendicular (right blue that bounces (is reversed): upper green.
Two green components add up to outgoung velocity (right red).

For your exercise () it might indeed be more sensible to let the mirror be at 22.5 degrees wrt horizontal. Make a similar drawing to convince yourself the outgoing velocity is then at 45 degrees.

And yes, in such a case you indeed get a considerable horizontal distance of travel !

More, in fact, than you calculated: in the y-direction the vertical velocty at impact is not zero but the negative of the initial velocity. As you can see from your own drawing.

It is a nice exercise to show that this 22.5 degrees indeed gives the worst-case range .