Kinematics

(a) The acceleration of the ball is \vec{a}=(1.20m/s^{2})\hat{\text{i}}-(9.8m/s^{2})\hat{\text{j}}.
Since \vec{a} is constant, the path is \vec{{\bf r }}=\vec{{\bf a}}t^{2}/2
,because \vec{{\bf v}}_{0}=0 and we can choose \vec{{\bf r}_{0}}=0.
This path is a straight line, with a direction given by \vec{{\bf a}}.
And then
\theta  =  arctan(9.80/1.20)=83.0^{o}.\\ R  =  \frac{39.0m}{tan(83.0^{o})}=4.79m.
It will be useful to find the height:
H=\frac{39.0m}{sin(83.0^{o})}=39.3m.


(b) The magnitude (size) of the acceleration of the ball is  a=\sqrt{9.80^{2}+1.20^{2}}(m/s^{2})=9.87m/s^{2}. The
time for the ball to travel down the hypotenuse of the figure is t=\sqrt{2(39.3m)/(9.87m/s^{2})}=2.82s.

(c) The magnitude of the speed of the ball at the bottom will be
v=at=(9.87\frac{m}{s^{2}})(2.82s)=27.8\frac{m}{s}.