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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.

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

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
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