# Homework Help: Finding distance traveled by rebound when measured for surface

1. May 16, 2017

### Ithilrandir

1. The problem statement, all variables and given/known data
Ball falling from 1m above a slanted surface 5 degrees from horizontal. KE conserved
Find distance travelled along slanted surface on its second rebound

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

3. 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

2. May 16, 2017

### 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 !)

3. May 16, 2017

### 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)

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4. May 16, 2017

### BvU

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 ?

5. May 16, 2017

### 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 explaination for the question.

6. May 16, 2017

### 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.

7. May 16, 2017

### Ithilrandir

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

8. May 16, 2017

### haruspex

sin 5°?

9. May 17, 2017

### BvU

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.

10. May 17, 2017

### haruspex

I was querying the trig function, not the angle.

11. May 17, 2017

### BvU

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

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

12. May 18, 2017

### Ithilrandir

I got the answer its apparent 0.68something meters

13. May 18, 2017

### ehild

Show please, how did you get that result.

14. May 18, 2017

### 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