Answer
a) $67.8m$
b) $7.38\frac{m}{s}$
Work Step by Step
a) $R=\frac{v^2\sin(2\theta)}{g}$
$R=\frac{\big(28\frac{m}{s}\big)^2\sin(2\times61^o)}{9.8\frac{m}{s^2}}=67.8m$
b) $y_f=y_i+v_{yi}t-\frac{1}{2}gt^2$
$0=0.90m+28\frac{m}{s}\times \sin(61)t-4.9\frac{m}{s^2}t^2$
$4.9t^2-24.5t-0.9=0$
$t=-0.036; t=5.03$
Time cannot be negative, so the second solution is correct.
The position of catching the ball
$x_c=105m-67.8m=37.2m$
$v_c=\frac{x_c}{t}=\frac{37.2m}{5.03s}=7.38\frac{m}{s}$
