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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!!!!!!.

Grade:12

1 Answers

Umang Agarwal
19 Points
8 years ago
Ux=10√3 m/s and Uy=30 m/s
x=Ux t + ½ Ax t2
t=x/Ux...(1)
Now, y=Uy t – ½ g t2
y=Uy x/Ux – ½ g (x/Ux)2
y=√3x – 1/60 x2...(2)
And given curve= y=b/a x
= y=1/√3 x...(3)
from (2) & (3),we get
y=3y – 1/20 y2...(4)
therefore intersection points by solving (4) are,
(0,0) & (40√3,40) where (40√3,40) is ans..

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