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CBSE Class 11-science Answered

1.A ball starts falling with 0 initial velocity on a smooth inclined plane forming an angle alpha with the horizontal.Having fallen the distance h,the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time?
Asked by smanishkumar2002 | 20 Apr, 2018, 09:59: PM
answered-by-expert Expert Answer
After travelling a height h, the ball which is started with zero initial velocity acquires the velocity u =
  As shown in figure, if the ball rebounds after hitting the inclined plane, then projection angle with respect to horizontal direction is (90-2α).
Horizontal velocity component = u×cos(90-2α) = u×sin(2α)
Vertical velocity component = u×sin(90-2α) = u×cos(2α)
 
maximimum vertical height H reached by the ball after rebounding :-   0 =  u2×cos2(2α) - 2gH ; hence H = [ u2×cos2(2α) ] / 2g
Time taken T1 to reach the height H is given by,  T1 = [ u×cos(2α) } /g ;
after reaching the height H, the ball will travel a distance H+L×tanα in a time T2 to reach the ground level. we get time T2 as given below
 
H+L×tanα = (1/2)g (T2)2 ; hence T2 = { (2/g) (H+L×tanα) }1/2  
 
Then horizontal Range = [ u×sin(2α) ](T1+T2)
(user is requested to do the required algebra to get the final answer )
Answered by Thiyagarajan K | 21 Apr, 2018, 10:36: PM
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