CBSE Class 11-science Answered
14?
Asked by smanishkumar2002 | 09 Jun, 2018, 07:55: AM
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,
..............................(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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