Home
Class 12
PHYSICS
Two projectile are projected simultaneou...

Two projectile are projected simultaneously from a point on the ground "O" and an elevated position "A" respectively as shown in the figure. If collision occurs at the point of return of two projectiles on the horizontal surface, then find the height of "A" above the ground and the angle at which the projectile "O" at the ground should be projected.

Text Solution

Verified by Experts

There is no initial separation between two projectile is x-direction. For collision to occur, the relative motion in x-direction should be zero. In other words, the component velocities in x-direction should be equal to that two projetiles cover equal horizontal distance at any given time. Hence,
`u_(Ox)=u(Ax)`
`impliesu_0costheta=u_A`
`impliescostheta=(u_A)/(u_0)=5/(10)=1/2=cos60^(@)`
`impliestheta=60^(@)`
We should ensure that collision does occur at the point of return. It means that by the time projectiles travel horizontal distances required, they should also cover vertical distances so that both projectile areat "C" at the same time. In the nutshell, their times of flight should be equal. For projectile from "O"
`T=(2u_Osintheta)/(g)`
For projectile from "A",
`T=sqrt((2H)/(g))`
For projectile from " A"
`T=(2u_0sintheta)/g=sqrt(((2H)/(g)))`
Squaring both sides and putting values,
`impliesH=(4u_(O)^(2)sin^2theta)/(2g)`
`impliesH=(4xx10^2sin260^(@))/(2xx10)`
`H=20((sqrt3)/(2))^2=15m`
We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding at the end of their flights and hence their times of flight should be equal. We can, however, proceed to analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles in the vertical direction is "H". This separation is covered with the component of relative in vertical direction.
`impliesv_(OAy)=u_(Oy)-u_(Ay)=u_Bsin60^(@)-0=10xx(sqrt3)/2`
`=5sqrt3m//s`
Now, time of flight of projectile from ground is :
`T=(2u_0sintheta)/(g)=(2xx10xxsin60^(@))/(10)=sqrt3`
Hence, the vertical displacement of projectile from "A" before collision is :
`impliesH=v_(OAy)XT=5sqrt3xsqrt3=15m//s`
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS

    MOTION|Exercise EXERCISE 1 OBJECTIVE PROBLEMS|103 Videos
  • KINEMATICS

    MOTION|Exercise EXERCISE 2 OBJECTIVE PROBLEMS|75 Videos
  • HYDROSTATIC, FLUID MECHANICS & VISCOSITY

    MOTION|Exercise EXERCISE -3 (SECTION-B) PREVIOUS YEAR PROBLEM|7 Videos
  • KINETIC THEORY OF GASES

    MOTION|Exercise EXERCISE-3 SECTION-B|16 Videos

Similar Questions

Explore conceptually related problems

Two particles are projected simultaneously from the level ground as shown figure. They may collide after a time :

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

Two projectiles are projected simultaneously from two towers as shwon in figure. If the projectiles collide in the air, then find the distance "s" between the towers.

The angle of projection at which the horizontal range and maximum height of projectile are equal is

Two particles P & Q are projected simultaneously from a point O on a level ground in the same vertical plane with the same speed in directions making angle of 30^(@) and 60^(@) respectively with the horizontal.

Which of the following is the graph between the height (h) of a projectile and time (t), when it is projected from the ground

The horizontal range is equal to two times maximum height of a projectile. The angle of projection ofthe projectile is:

A projectile is hurled into air from a point on the horizontal ground at an angle with the vertical. If the air exerts a constat resistve force,

Particle is projected from the ground at a certain angle with horizontal . If particle takes time T to hit the ground again , then maximum height of the projectile is