Answer
Please see the work below.
Work Step by Step
(a) We know that the horizontal component of force is given as
$F_h=\frac{mv^2}{r}$
We plug in the known values to obtain:
$F_h=\frac{(45)(6.3)^2}{5.0}=357.21N$
The vertical component of the force is given as
$F_v=mg$
We plug in the known values to obtain:
$F_v=(45)(9.8)=441N$
(b) The required angle is given as
$\theta=tan^{-1}(\frac{F_v}{F_h})$
$\theta=tan^{-1}(\frac{441}{357.21})=51^{\circ}$
