Answer
(a) Speed of the dog is, $v_{dog}=8.5 m/s$
(b) Horizontal distance travelled by the ball, $R \approx13.3 m$
Work Step by Step
(a) For dog to catch the ball, he must has same speed as horizontal speed of the ball,
So, speed of the dog is, $v_{dog}=v_{x, ball}=8.5 m/s$
(b) Since ball is projected horizontally, the time of flight for the ball will be,
$T=\sqrt \frac{2H}{g}=\sqrt \frac{2\times 12}{9.8}\approx1.56 sec$
Range i.e. horizontal distance travelled by the ball,
$R=v_x T=v_x \sqrt \frac{2H}{g} = 8.5\sqrt \frac{2\times 12}{9.8} \approx13.3 m$
