A particle is projected from the bottom of an inclined plane of inclination 30 degrees with velocity of 40m/s at an angle of 60 degrees with thw horizontal.Find the speed of the particle when its velocity vector is parallel to the plane.

ANS GIVEN:40/underroot3

Dear Sameer,

let the plane be in the direction of x axis and the perpendicular to the inclined be y axis, then
when the velocity is in x direction only, then y component=0
So time=4/(root3) seconds.    [since v(y)=u-at...... u=40 sin30 ....a=gcos30 ]
speed is only in the x axis=v-(g/2)t=20(root3) - 5.4/(root3)
                                         =60-20 / root3
                                         =40/(root3).(ANS)

Regards.

its difficult to explain here but i''ll try. 1st thing u need 2 knw is how to rite components of eny given vector. if u knw ten follow the following steps

1. draw the diagram of wats happening. this will give an idea for the next step

2. now the method is tat v r using motion in 2-d. so v r going to take vector components. NOT along the usual but along the slope of the inclined plane,(rotation of axis with 30 degrees)  this means u have to take the velocity as 2 components, u(cos30)i+ u(sin 30)j. also take components of g along the plane wich is g(cos60)(-i) + g (sin 60)(-j) now u find the velocity of the particle at any given time.

3. mow the prob is like a motion in 2d and then equate the velocity in the j direction to 0 and put the value in the other equation.

Thank you!It helped me a lot.......

always ready to help

let the plane be in the direction of x axis and the perpendicular to the inclined be y axis, then velocity is
 in x direction only, then y component=0
So time=4/(root3) seconds.    [since v(y)=u-at...... u=40 sin30 ....a=gcos30 ]
speed is only in the x axis=v-(g/2)t=20(root3) - 5.4/(root3)
                                         =60-20 / root3
                                         =40/(root3).(ANS)