91影视

The velocity contains two components, one is horizontal component and another one is vertical.

According to the newton鈥檚 laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time and acceleration.

Vertical distance of the boy from the ground is 12.0m.

The speed of the ball is 8.50m/s.

(a)

The boy will have to travel with the same speed to reach the same horizontal distance as ball, so the speed also be the same that is 8.50m/s.

(b)

For the vertical motion, initial velocity is zero

$s=ut+\frac{1}{2}a{t}^2$

Substitute 12m for s, 0m/s foru and$9.8\mathrm{m}/{\mathrm{s}}^2$ forg in the above equation.

$$\begin{gathered} 12=0脳t+\frac{1}{2}脳9.8t^2 \\ t=1.56s \end{gathered}$$

For the horizontal motion, gravity is zero

$s=ut+\frac{1}{2}a{t}^2$

Substitute 8.50 s for u, 1.56s for t,$0m/s^2$ forg in the above equation.

$$\begin{gathered} s=8.50脳1.56+\frac{1}{2}脳0脳{\left(1.56\right)}^2 \\ s=13.3m \end{gathered}$$

Therefore the horizontal distance travel by the dog is 13.3 m.