Answer
The tension in the rope is $~~9.4~N$
Work Step by Step
We can find the angle $\theta$ that the rope makes with the vertical:
$tan~\theta = \frac{4.2~cm}{8.0~cm}$
$\theta = tan^{-1}~(\frac{4.2~cm}{8.0~cm})$
$\theta = 27.7^{\circ}$
The vertical component in the rope's tension $F_T$ is equal in magnitude to the ball's weight. We can find $F_T$:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
$F_T = \frac{(0.85~kg)(9.8~m/s^2)}{cos~27.7^{\circ}}$
$F_T = 9.4~N$
The tension in the rope is $~~9.4~N$
