61P - University Physics with Modern Physics

Question

Suppose that the boy in Problem 3.60 throws the ball upward at 60^o above the horizontal, but all else is the same. Repeat parts (a) and  (b) of that problem.

 

Problem 3.60: 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 8.50 m/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

1. The speed of the dog will be 4.25 m/s.
2. The horizontal distance traveled by the dog is 10.54 m.

Step 1: Introduction:

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

 

According to Newton’s laws of motion,

 

$\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ 

 

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

Step 2: Consider the given data:

The vertical distance, s=12.0 m 

 

The horizontal velocity of the ball, u=8.5 m/s

 

Angle, $\mathrm{\theta }={60}^{\mathrm{o}}$ 

Step 3: (a) The speed of a dog:

The horizontal and vertical velocity components are   and   respectively. So the horizontal and vertical velocity will be

 

$\begin{array}{rcl}\mathrm{u} \mathrm{cos\theta }& =& 8.5 \mathrm{m}/\mathrm{s}\times \cos {60}^{\mathrm{o}}\\ & =& 4.25 \mathrm{m}/\mathrm{s} \end{array}$ 

$\begin{array}{rcl}\mathrm{u} \mathrm{sin\theta }& =& 8.5 \mathrm{m}/\mathrm{s}\times \sin {60}^{\mathrm{o}}\\ & =& 7.36 \mathrm{m}/\mathrm{s}\end{array}$ 

                                                                      

The boy will have to travel at the same speed to reach the same horizontal distance as the ball, so the speed also be the same that is 4.25 m/s.

Step 4: (b) Distance of the ball from the tree:

For the vertical motion, the initial velocity is zero.

 

 $\begin{array}{rcl}\mathrm{s}& =& \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ 12 \mathrm{m}& =& \left(7.36 \mathrm{m}/\mathrm{s}\times \mathrm{t}\right)+\frac{1}{2}\left(9.8 \mathrm{m}/{\mathrm{s}}^2\times {\mathrm{t}}^2\right)\\ \mathrm{t}& =& 2.48 \mathrm{s}\end{array}$

 

For the horizontal motion, gravity is zero

                                                                       

$\begin{array}{rcl}\mathrm{s}& =& \mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2\\ & =& 4.25 \mathrm{m}/\mathrm{s}\times 2.48\mathrm{s}+\frac{1}{2}\times 0 \mathrm{m}/{\mathrm{s}}^2\times {\left(2.49 \mathrm{s}\right)}^2\\ & =& 10.54 \mathrm{m}\end{array}$    

 

Hence, the horizontal distance traveled by the dog is 10.54 m.