Q60P - University Physics with Modern Physics

Question

A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?

Step-by-Step Solution

Answer

(a) The speed of the dog will be 8.50m/s.

(b) The horizontal distance travel by the dog is 13.3m. 

Step 1: Introduction

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

 

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$ 

 

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

Step 2: Given data

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

 

The speed of the ball is 8.50m/s.

Step 3: The speed of dog

(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.

Step 4: Distance of the ball from the tree

(b)

 

For the vertical motion, initial velocity is zero

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

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

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

 

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, $0m/s^2$ for g in the above equation.

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

 

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