Answer
The maximum angle with the vertical that the rope can make is $~~19^{\circ}$
Work Step by Step
In part (a), we found that the child weighs is $~~270~N$
The vertical component of the tension is equal in magnitude to the child's weight.
The force exerted by the father, which is $93~N$, is equal in magnitude to the horizontal component of tension.
We can find the maximum angle with the vertical that the rope can make:
$tan~\theta = \frac{93~N}{270~N}$
$\theta = tan^{-1}~(\frac{93~N}{270~N})$
$\theta = 19^{\circ}$
The maximum angle with the vertical that the rope can make is $~~19^{\circ}$
