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.

http://katus.kabel1.cz/homework.jpg [Broken]

The ball falls down on the incline from height

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.

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]

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?

**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?

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