Home
Class 11
PHYSICS
Determine the horizontal velocity v0 wit...

Determine the horizontal velocity `v_0` with which a stone must be projected horizontally from a point P, so that it may hit the inclined plane perpendicularly. The inclination of the plane with the horizontal is `theta` and point P is at a height h above the foot of the incline, as shown in the figure.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C

`u_x = v_0 cos theta, u_y = v_0sin theta, a_x = -g sin theta`,
`a_y = g cos theta`
At Q, `v_x = 0`
`:. u_x + a_xt = 0`

or `t = (v_0cos theta)/(g sin theta)` ……..(i)
`s_y = h cos theta`
`:. u_yt + 1/2 a_yt^2 = hcos theta`
`:. (v_0 sin theta)((v_0cos theta)/(g sin theta))`
`+ 1/2 (g cos theta)((v_0 cos theta)/(g sin theta)) = h cos theta`
Solving this equation we get,
`v_0 = sqrt(2gh)/(2+cot^2theta)`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • PROJECTILE MOTION

    DC PANDEY ENGLISH|Exercise Level - 2 More Than One Correct|8 Videos
  • MOTION IN A PLANE

    DC PANDEY ENGLISH|Exercise (C )Medical entrances gallery|32 Videos
  • PROPERTIES OF MATTER

    DC PANDEY ENGLISH|Exercise Integer|8 Videos

Similar Questions

Explore conceptually related problems

At what angle should a ball be projected up an inclined plane with a velocity v_0 so that it may hit the incline normally. The angle of the inclined plane with the horizontal is prop . .

A stone must be projected horizontally from a point P, which is h meter above the foot of a plane inclined at an angle theta with horizontal as shown in figure . Calculate the velocity v of the stone so that in may hit the incline plane perpendiculary.

In figure, the angle of inclination of the inclined plane is 30^@ . Find the horizontal velocity V_0 so that the particle hits the inclined plane perpendicularly. .

A particle of mass m is thrown horizontally from point P , as shown in the figure, with speed 5 m//s . It strikes an inclined plane perpendicularly as shown. The inclination of the plane is 30^(@) with the horizontal. Then, the height h shown in the figure is ( Take g = 10 m//s^(2) ) .

The velocity of a sphere rolling down an inclined plane of height h at an inclination theta with the horizontal, will be :

A projectile si thrown with velocity of 50m//s towards an inclined plane from ground such that is strikes the inclined plane perpendiclularly. The angle of projection of the projectile is 53^(@) with the horizontal and the inclined plane is inclined at an angle of 45^(@) to the horizonta.. (a) Find the time of flight. (b) Find the distance between the point of projection and the foot of inclined plane.

A ball is projected horizontally from an inclined plane with a velocity V, as shown in the figure. It will strike the plane after a time

A particle is projected horizontally with spee u from point A, which is 10 above the ground . If the particle hits the inclined perpendicularly at point B. [ g = 10 m//s^(2)] Find the length OB along the inclined plane

A ball is projected horizontally with a speed v from the top of the plane inclined at an angle 45^(@) with the horizontal. How far from the point of projection will the ball strikes the plane?

A particle is projected horizontally with spee u from point A, which is 10 above the ground . If the particle hits the inclined perpendicularly at point B. [ g = 10 m//s^(2)] Speed at point B will be