Answer
$r_p=3.15m$
Work Step by Step
We can find the required distance as follows:
$r_p=\frac{LTsin\theta-\frac{1}{2}Lm_rg}{m_pg}$
We plug in the known values to obtain:
$r_p=\frac{(4.25)(1450)sin30.0^{\circ}-\frac{1}{2}(4.25)(9.81)}{(68.0)(9.81)}$
$r_p=3.15m$
