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CBSE Class 11-science Answered

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Asked by smanishkumar2002 | 09 Jun, 2018, 07:55: AM
answered-by-expert Expert Answer
 
Let the ball is dropped from point O as shown in figure.
Let h be the distance travelled by the ball vertically just before hitting the inclined plane at A.
 
velocity u of the ball when it reaches the point A is given by, 
 
begin mathsize 12px style u space equals space square root of 2 g h end root end style ..............................(1)
After hitting the point A, the ball rebounds due to elastic collision and
its velocity vector at point A makes an angle α with the vertical line drawn on the inclined plane as shown in figure. 
 
From this point A onwards, we can consider the motion of the ball as a projectile motion on the inclined plane
with initial velocity u as given in equation (1) and projected at an angle α with the vertical line drawn on inclined plane.
 
vertical component of velocity of the ball with respect to the inclined plane is u×cosα
horizontal component of velocity of the ball with respect to the inclined plane is u×sinα
 
Acceleration due to gravity also resolved into two components,
one parallel to inclined plane and other perpendicular to incline dplane as shown in figure
 
As in projectile motion, let B be the point of maximum height on inclined plane
so that the vertical component of velocity becomes zero at B.
Time tB taken by the ball to travel from A to B is given by,  tB = ucosα/(gcosα) = u/g
Time tC taken by the ball to travel from A to C is given by,  tC = 2u/g
during this time tC, distance AC travelled parallel to inclined plane is given by
 
AC = usinα × (2u/g) +(1/2)×gsinα×(4u2/g2).............(2)
 
( equation of motion used to get (2) is  " s = ut+(1/2)at2 "  )
 
By substituting for u using eqn.(1) in eqn(2) and after simplification, we get AC = 8h×sinα
Answered by Thiyagarajan K | 10 Jun, 2018, 10:44: AM
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