A stone is projected from a horizontal plane. It attains height 'H' & strikes a stationary smooth wall & falls on the ground vertically below the maximum height. Assume the collision to be ellastic, find the height of the point on the wall where ball will strike.

Dear Akash,

let 'v' be speed of projection and theta be angle.

H=(v*sin(theta))²/2g,range=R=v²sin(2*theta)/g,R/H=cot(theta)/g,R=(Hcot(theta)/g)

let 'h' be answer,

let alpha be angle with normal to wall, both just before and just after collision (collision is elastic)

magnitude of vertical component of velocity at instant of impact=√(v²sin²(theta)-2gh)=,horizontal distance between point of projection and wall be 'x',x=(R/2)+(R/4)=3R/4=(3Hcot(theta)/4g)

(x-(R/2))/vcos(theta)=R/4vcos(theta)=vsin(theta)/2g=√(H/2g)=time between wall impact and hitting ground=T

h=uT+(g/2)T², vsin(theta)-u=gT,total time of flight=2vsin(theta)/g,T=(2vsin(theta)-u)/g=(2√(2gH)-√(2g(H-h)))/g

answer=h=uT+(g/2)T²,u=√(2g(H-h)),T=(2√(2gH)-√(2g(H-h)))/g,Solve for 'h'

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Approved

Get R equal to 4Hcot reheta by deciding R and H..The wall will be 3R/4 away from projection point ..Use equation of path of projectile y= xtan theeta( 1-x/R)..Put x equals to 3R/4 and R=4H cot theeta..... Ans 3H/4