Answer
(a) Angle of projection, $\theta=72.5^0$
Horizontal distance travelled by the women, $x=33.3m$
(b)
Work Step by Step
(a) For women to catch the ball, she must have same speed as horizontal speed of the ball i.e. $v_w=v_{x,b}=ucos\theta \Rightarrow cos\theta=\frac{v_w}{u}=\frac{6}{20}=0.3 \Rightarrow \theta=72.5^0$
Now to find time of flight for the ball, we will use $2^{nd}$ equation of motion in vertical direction,
$-h=usin\theta t-\frac{1}{2}gt^2 \Rightarrow 4.9t^2-19.1t-45=0$
Positive solution of this equation is, $T=5.55 s$
So horizontal distance travelled by the women, $x=v_wT=6\times 5.55=33.3m$
(b) To a observer on ground, ball will appear to be moving on a parabolic path and to the runner, since there is no relative motion in horizontal, so ball will appear to be moving in vertical direction only.
