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an object is projected from origin with initial speed 20√3 m/s and angle 60 degree with x axis.Taking ground as x axis an inclined plane is kept on the line ay=bx..Calculate the coordinates of the point where it meets the inclined plane,where a=√3,b=1..I am not able to solve this .plz explain!!!!!!.
an object is projected from origin with initial speed 20√3 m/s and angle 60 degree with x axis.Taking ground as x axis an inclined plane is kept on the line ay=bx..Calculate the coordinates of the point where it meets the inclined plane,where a=√3,b=1..I am not able to solve this .plz explain!!!!!!.

```
5 years ago

Umang Agarwal
19 Points
```							Ux=10√3 m/s and Uy=30 m/sx=Ux t + ½ Ax t2t=x/Ux...(1)Now, y=Uy t – ½ g t2y=Uy x/Ux – ½ g (x/Ux)2y=√3x – 1/60 x2...(2)And given curve= y=b/a x= y=1/√3 x...(3)from (2) & (3),we gety=3y – 1/20 y2...(4)therefore intersection points by solving (4) are,(0,0) & (40√3,40) where (40√3,40) is ans..
```
5 years ago
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