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.