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.

Chetan Mandayam Nayakar
312 Points
11 years ago

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

Hello moderator:in case you need any clarification, please contact me at mn_chetan@yahoo.com

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Regards,

CHETAN MANDAYAM NAYAKAR

Mohan
11 Points
5 years ago
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
Samuel Garry
61 Points
4 years ago

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. ........ it is also obvious by seeing the diagram that height of wall will be 3x by 4