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Ball Motion Down the Incline

  • Thread starter Kat3rina
  • Start date
Hi. I have problem with my homework. I have already searched related theoretic basis and I have already tried to solve it too, but I need some feedback, if my solution is right. I would like to ask you for it. Thank you very much.

1. Homework Statement
http://katus.kabel1.cz/homework.jpg [Broken]
The ball falls down on the incline from height h (known variable). Then it rebounds according to the law of reflection (angle δ). The ball has to fly through the hole in the wall standing in distance x (known variable). Angle of the incline is α.
Given data:
[tex]h = 0,5 m[/tex]
[tex]x = 0,15 m[/tex]
[tex]\alpha = 15°[/tex]
and all the variables in the picture typed bold.

2+3. Relevant equations and the attempt at a solution
Free fall
acceleration:
[tex]a = g[/tex]
velocity:
[tex]v = \int a \ dt = g t + v_0 [/tex]
height:
[tex]h = \int v \ dt = \frac{1}{2} g t^2 + v_0 t [/tex]
[tex]v_0 = 0[/tex]
time of impact:
[tex]t = \sqrt {\frac{2h}{g}}[/tex]
velocity in the time of impact:
[tex]v = g t = \sqrt {2gh}[/tex]

Oblique throw
accelerations:
[tex]a_x = 0[/tex]
[tex]a_y = -g[/tex]
velocities:
[tex]v_0 = \sqrt {2gh}[/tex]
[tex]v_x = \int a_x \ dt = v_0 cos \beta [/tex]
[tex]v_y = \int a_y \ dt = -g t + v_0 sin \beta [/tex]
distances:
[tex]x = \int v_x \ dt = v_0 cos \beta t + x_0[/tex]
[tex]y = \int v_y \ dt = - \frac{1}{2} g t^2 + v_0 sin \beta t + y_0[/tex]
[tex]x_0 = 0[/tex]
[tex]y_0 = 0 [/tex]
time of contact with the wall:
[tex]t = \frac{x}{v_0 cos \beta}[/tex]
[tex]y = x tg \beta - \frac{x^2}{4 h (cos \beta)^2}[/tex]

My results with the given data:
[tex]y = 0,257 m[/tex]
Is it right, please?
 
Last edited by a moderator:

LowlyPion

Homework Helper
3,079
4
Welcome to PF.

It seems a little complicated to not solve for the intermediate results along the way, but your answer looks in close agreement to what I get.

I think your last statement for y is not quite right, however.
 
Thank you very much for your help!

Do you think the last y statement is wrong?

I get it from the previous statements this way:

[tex]x = v_0 cos \beta \cdot t[/tex]
[tex]t = \frac {x}{v_0 cos \beta}[/tex]
[tex]y = - \frac {1}{2}g t^2 + v_0 sin \beta \cdot t[/tex]
[tex]y = - \frac {1}{2}g (\frac {x}{v_0 cos \beta})^2 + v_0 sin \beta \cdot {\frac {x}{v_0 cos \beta}}[/tex]
[tex]y = - \frac {1}{2}g \frac {x^2}{v_0^2 (cos \beta)^2} + sin \beta \cdot {\frac {x}{ cos \beta}}[/tex]
[tex]v_0 = \sqrt{2gh}[/tex]
[tex]y = - \frac {1}{2}g \frac {x^2}{2gh (cos \beta)^2} + sin \beta \cdot {\frac {x}{cos \beta}}[/tex]
[tex]y = - \frac {1}{4} \frac {x^2}{h (cos \beta)^2} + x \cdot {\frac {sin \beta}{cos \beta}}[/tex]
[tex]y = - \frac {x^2}{4 h (cos \beta)^2} + x \cdot tg \beta[/tex]
 

LowlyPion

Homework Helper
3,079
4
Thank you very much for your help!

Do you think the last y statement is wrong?

I get it from the previous statements this way:

[tex]x = v_0 cos \beta \cdot t[/tex]
[tex]t = \frac {x}{v_0 cos \beta}[/tex]
[tex]y = - \frac {1}{2}g t^2 + v_0 sin \beta \cdot t[/tex]
[tex]y = - \frac {1}{2}g (\frac {x}{v_0 cos \beta})^2 + v_0 sin \beta \cdot {\frac {x}{v_0 cos \beta}}[/tex]
[tex]y = - \frac {1}{2}g \frac {x^2}{v_0^2 (cos \beta)^2} + sin \beta \cdot {\frac {x}{ cos \beta}}[/tex]
[tex]v_0 = \sqrt{2gh}[/tex]
[tex]y = - \frac {1}{2}g \frac {x^2}{2gh (cos \beta)^2} + sin \beta \cdot {\frac {x}{cos \beta}}[/tex]
[tex]y = - \frac {1}{4} \frac {x^2}{h (cos \beta)^2} + x \cdot {\frac {sin \beta}{cos \beta}}[/tex]
[tex]y = - \frac {x^2}{4 h (cos \beta)^2} + x \cdot tg \beta[/tex]
Sorry. I didn't recognize your tgβ notation as tanβ with t and g also being variables of the solution.
 

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