Ball Motion Down the Incline: Is the Calculated Height Correct?

[Kat3rina]

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.

Homework Statement

http://katus.kabel1.cz/homework.jpg
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:
$$h = 0,5 m$$
$$x = 0,15 m$$
$$\alpha = 15°$$
and all the variables in the picture typed bold.

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

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

My results with the given data:
$$y = 0,257 m$$
Is it right, please?

 

[LowlyPion]

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.

 

[Kat3rina]

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:

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

 

[LowlyPion]

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:

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

Sorry. I didn't recognize your tgβ notation as tanβ with t and g also being variables of the solution.