### 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
$v=at=(9.87\frac{m}{s^{2}})(2.82s)=27.8\frac{m}{s}.$