yeah the answer is here...
On falling a distance h vertically the velocity of the body is v = <sq. root>(2gh) ----->(from mgh = 1/2 mv^2)
Then on bouncing elastically the body regains its initial velocity in the direction parallel to horizontal (i,e, at an angle <alpha> to the normal N) as shown in figure.
resolve this velocity as shown.
Let the body falls a distance l along the incline before hitting the incline again.
So taking x- axis as the inclined plane and y-axis as the normal N we get that the body has no displacement along y-axis.
let t = time for which the body remains in air after first impact.
So 0 = vcos<alpha>.t - 1/2 g.cos<alpha>.t^2 ----> (as g.cos<alpha> is directly opposing motion along y- axis and produces retardation)
or t = 2v/g
or t = (2/g).<sq.root>(2gh) -----> (as v = <sq. root>(2gh) )
Then considering motion along x - axis we find that unlike normal projectile sums there is a accl^n along the x-axis too.
So in this case
l = vsin<alpha>.t + 1/2 gsin.<alpha>.t^2 -----> (as g.sin<alpha> is along the direction of motion and causes accl^n)
= <sq.root>(2gh).sin<alpha>.(2/g).<sq.root>(2gh) + 1/2 g.sin<alpha>.[(2/g).<sq.root>(2gh)]^2
= 2gh.(2/g).sin<alpha> + (1/2).g.sin<alpha>.[4/(g^2)].2gh
= 4h.sin<alpha> + (2gh).(g/2).[4/(g^2)].sin<alpha>
= 4h.sin<alpha> + (g^2)h.[4/(g^2)].sin<alpha>
= 4h.sin<alpha> + 4h.sin<alpha>
= 8h.sin<alpha>
solved...
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
0
with the horizontal.Having fallen the distanceH,the ball rebonds elastically off the inclined plane.At what distance from the impact pointwill the ball reboundfor the second time? 
1
![[Thumb - irodov.JPG]](/upload/2007/6/29/dcf59082ee5c257ebcd5699f02274912_19520.JPG_thumb)
is 45 degrees then only the body moves horizontal otherwise not.
