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# 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
9 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 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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We

All the best Akash !!!

Regards,

CHETAN MANDAYAM NAYAKAR

Mohan
11 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
Samuel Garry
61 Points
3 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