Given :
h= 1.57 m, x = 3.17 m, and vb = 3.03 m/s .
To Find :
The upward velocity [tex]v_b[/tex] of cylinder .
Solution :
Component of [tex]v_a[/tex] along string [tex]v_s= v_acos\ \theta[/tex] .
Now ,
[tex]tan\ \theta=\dfrac{h}{x}\\\\cos\ \theta=\dfrac{x}{\sqrt{x^2+h^2}}[/tex]
Now , from the figure :
[tex]2v_b=v_s\\\\2v_b=v_acos\ \theta\\\\v _a=\dfrac{2v_b}{cos\ \theta}\\\\v_a=\dfrac{2v_b\sqrt{x^2+h^2}}{x}[/tex]
Putting all given values in above equation :
[tex]v_a=\dfrac{2\times 3.03\times \sqrt{3.17^2+1.57^2}}{3.17}\\\\v_a=6.76\ m/s[/tex]
Therefore , value of [tex]v_a[/tex] is 6.76 m/s .
Hence , this is the required solution .
