Finding distance traveled by rebound when measured for surface

[Ithilrandir]

Homework Statement

Ball falling from 1m above a slanted surface 5 degrees from horizontal. KE conserved
Find distance traveled along slanted surface on its second rebound

Homework Equations

S=(1/2)(at^2)
(V^2)=(u^2)-2as

The Attempt at a Solution

V before hitting =sqrt(2*9.81)
=4.42
4.42/9.81=0.4515
0.4515*2=0.9030
0.9030*4.42*cos(85 degrees)
=0.0355
0.0355/cos(85)=3.98

 

[BvU]

Hello Itilrandir,

I can follow you up to ##v=4.42## m/s (always include units). Could you explain what you are doing after that ? And make a sketch of the situation and post it ?

I do notice there is no question in your post ?

(all in the spirit of PF culture, you mostly don't get a yes/no answer and we don't do the exercise for you. But we help where we can !)

 

[Ithilrandir]

Sorry for not being very clear with myself.
The question basically asks for us to find distance between the first and second impact on a infinitely long slanted surface tilted at 5 degrees.
Pic attached is what I attempted to do and get the final ans of around 4(3.98)

 

[BvU]

Could you explain what you are doing after that ?

I do understand the question.

Let me guess (this is NOT PF culture -- instead, you should explain, as I asked ) :
You use ##v = v_0 - gt = 0 \Rightarrow t = {v_0\over g} ## to find the time to reach the highest point after the bounce, 0.903 m.

However, ##\vec v_0## is not in the same direction as ##g## is acting, so you lose me at that point.
Furthermore, I also don't understand (understatement) the 85 degrees in the next line...

In your sketch I see 1 m above the surface as a constant -- which it may well be. Can you prove that it is a constant ?

 

[Ithilrandir]

It was just a random guess/ attempt at solving it cos the question was timed when i was doing it. I was hoping that someone has a better explanation for the question.

 

[BvU]

After the first bounce, ##v=v_0=4.42## m/s, I agree.
Can only be if it is at 10##^\circ## wrt the vertical (##E=\vec p^2/(2m)## is conserved).

So you have ##v_x = v_0\cos(80^\circ),\ v_y = v_0\sin(80^\circ),\ ## for a parabolic trajectory.
That intersects the surface ##y = y_0 - x\sin 5^\circ## at the second bounce point.
Set up the equations and solve the quadratic equation.

 

[Ithilrandir]

Do you have the correct answer? I want to double check after doing it. No time right now.

 

[haruspex]

After the first bounce, ##v=v_0=4.42## m/s, I agree.
Can only be if it is at 10##^\circ## wrt the vertical (##E=\vec p^2/(2m)## is conserved).

So you have ##v_x = v_0\cos(80^\circ),\ v_y = v_0\sin(80^\circ),\ ## for a parabolic trajectory.
That intersects the surface ##y = y_0 - x\sin 5^\circ## at the second bounce point.
Set up the equations and solve the quadratic equation.

sin 5°?

 

[BvU]

sin 5°?

a slanted surface 5 degrees from horizontal

Do you have the correct answer? I want to double check after doing it. No time right now.

PF culture is YOU do the work, WE try to help . If that necessitates working it all out, then I am always willing to do the work. Not in this case.

 

[haruspex]

a slanted surface 5 degrees from horizontal.

I was querying the trig function, not the angle.

 

[BvU]

With good reason . Should have been ##\tan## ! Thanks!

Now we both wait for Itil to have some time to work it out ...

 

[Ithilrandir]

I got the answer its apparent 0.68something meters

 

[ehild]

I got the answer its apparent 0.68something meters

Show please, how did you get that result.

 

[Ithilrandir]

Sorry I was busy.
The method to solve it is to resolve the components of gravity and velocity and turn it into a projectile motion