Answer
(a) Speed of the dog is, $v_{dog}=4.25 m/s$
(b) Horizontal distance travelled by the ball, $R \approx10.58 m$
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, $v_{dog}=v_{x, ball}=8.5\times cos60^0=4.25 m/s$
(b) Using $2^{nd}$ equation of motion in vertical direction (taken upward direction as positive),
$s_y=u_y t+\frac{1}{2}a_yt^2\Rightarrow -12=8.5sin60^0 t+\frac{1}{2}(-9.8)t^2$
$\Rightarrow 4.9t^2-7.36t-12=0$
Positive solution of this quadratic equation is, $T\approx 2.49 sec$
So, range i.e. horizontal distance travelled by the ball will be,
$R=v_x T= = 4.25\times2.49 \approx10.58 m$
