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  1. Oct 12, 2007 #1

    small elastic ball is dropped from the height h = 1.1 m onto an incline. After hitting the incline the ball bounces and hits the incline the second time. The ball bounces at same speed with which it hits the incline. The angle of the incline is 32 degrees. The angle between the normal to the incline and the velocity before the first collision is the same as the angle between the normal to the incline and the velocity after the first collision ( like a reflection of a light beam in geometrical optics) . Calculate the distance between the first and the second impact.
  2. jcsd
  3. Oct 12, 2007 #2


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    You need to show some work to get help. What did you try?
  4. Oct 12, 2007 #3
    What I 've know about this question's information is to find the distace between position
    of first hit and second hit of droping ball from 1.1 m height. Also it is I thinki have to use this formular; y=x*tan(32)-9.8*x^^2/2*(v0*cos(theta)^^2) is that right?
    answer is 4.7m but I really don;t understand I 've been trying for few hours for this pleez
  5. Oct 13, 2007 #4


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    Where did you get that formula? Did you work it out for yourself? I don't think it is quite correct. Can you specify what is giving you the most trouble?

    Can you figure out the velocity that the ball hits the incline with?

    Have you drawn a diagram and figured out the way the ball bounces?

    You can use [tex]h = h_0 + v_{0y }t + (1/2) g t^2[/tex] and [tex]v_{x} = \Delta x / \Delta t[/tex] where x is the horizontal distance between the two impact points, NOT the distance along the incline between the two impact points (which is what you want). You need to find a way to relate those two distances.
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