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`        A particle is projected horizontally with speed u from point A ,which is 10 m above the ground .If the particle hits the inclined plane perpendicularly at pointbl B .please give me detailed solution.`
3 months ago Saurabh Koranglekar
3296 Points
```							Dear studentPlease find the link belowhttps://www.askiitians.com/forums/Mechanics/a-particle-is-projected-horizontally-with-speed-u_152557.htmRegards
```
3 months ago
```							Dear student Taking positive x axis parallel to OB and y axis perpendicular to it (downward as positive) Initial conditionunder initial conditionux = ucos 45uy = usin 45o ax = -gsin45ay = gcos45to determine tusing equation of motion along x axisvx = ux +axtas vx = 0 so t = -(ux/ax)or, t =  ucos 45/gsin45​t = u/gthis is the time of flightnow using itwe drop a perpendicular from projection point on OB​the length of the projection is l = 10 sin45​ using equation of motion along y axisy = uy + ayt^2/210sin 45 = usin45 + 12gcos 45(u/g)2we get 10 = u + u^2/2gor u^2 + 2ug = 20gor, u^2 +20u - 200 = 0solving this quadratic equation we get,u = 7.32 m/s ​ at point B ,  it has only velocity in y direction so its velocity in x direction has become zerohence using the equation of motion along y axisvx^2 = ux^2 + 2axas vx = 0so x = -(u cos45)2 / 2(-gsin 45)x = u^2 /2sqrt(2)x = 1.89 m so using equation of motion along y axisvy = uy +aytvy = usin 45 + u(gcos45)/gvy = usin45 + ucos45thus total velocity at B is vy as vx = 0v = 7.32 * sqrt(2)  = 10.35 m /s
```
3 months ago
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