IIT JEE , AIEEE Forum - Physics , Mechanics

raulrag009

It's an irodov question!!!!

THE ball will hit the plane with some velocity and rebound with the same velocity , follow a projectile path and hit the plane again

since the ball has travelled for height 'h'

it's velocity while hittin is

v²=0+2gh

v=

2gh

since the collision is elastic

the ball will rebound with the same velocity

It's initial velocities are

In X axis In Yaxis

u_x = -vsinx u_y= vcosx

a_x= -gsinx a_y= -gcosx

now on the inclined plane,after falling for the second time

It's displacement in y-axis will be zero {w.r.t inclined plane}

In Y axis

0=vcosxt - (1/2)gcosxt^2

t=2v/g

NOw

In X axis

Let R be the distance travelled before hitting on the inclined plane

R= -vsinxt - (1/2)dsinxt^2

put value of t

R = -2v^2sinx/g -4v^2sinx/2g

R = -8v^2sinx/2g

R = -8 (2gh)sinx/2g

R = -8hsinx

thus

mag(R)=8hsinx