I had 88.8 degrees but that is not the right answer to this problem. Please help.
length of fishing line 0.23 m is hypoteneus of right angelled triangle in vertical plane with base equal to radius 'r' of circle
Suppose in the vector triangle of forces, fishing line length Lrepresents tension in the line
If height 'h' of triangle represents weight 'mg' of sinker and base 'r' of triangle represents centripetal force F=mv^/r=mr*4*(pi)^2 / T^2
where T is time period =0.60 s
mg/h=mr4(pi)^2/rT^2
h=gT^2/4(pi^2)=9.8*0.6*0.6 / 4*9.87= 0.089 m
suppose fishing line makes angle O with vertical ,then
cosO=h/ L( lenght of line)
cosO=0.089/0.23=0.3885
O=67.13 degree
Angle O with vertical is 67.13 degree
It wouldn't be vertical. It would be "diagonal." and that is why
I would never have you in a boat with me, because, I would
end up with that hook in my eye.
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