ATHARVA_12
Last Activity: 4 Years ago
Let Vertex be (0,0) and let the parabola be
x2=4ay..... so by symmetry when x=10m then y=4m(maximum saag)
then,on substituting x and y we get
100=16a so 4a=25
hence,the eq of parabola is x2=25y as(4a=25.)
then at distance x=5 then let point on parabola be(5,w) then
52=25y …... so y=1
Now, as the sag is 4m and height of towers is 15m,So the the lowaset point of the wire is 11m above the ground
hence,the height of the point 5m away from one end is..
11+y-->11+1=>12m
so height of the point at a distance 5m from one end is 12m.