Problem 53 - College Physics

Question

\(\bullet\) 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 \(\mathrm{m} / \mathrm{s}\) , (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) 8.50 m/s, (b) 13.26 m

Step 1: Understanding the Problem

We have a boy in a tree at 12 m height who throws a ball horizontally with a speed of 8.50 m/s. We need to calculate the speed at which the dog must run to catch the ball at the ground level and the horizontal distance from the tree where the ball lands.

Step 2: Find the Time for the Ball to Hit the Ground

Since the ball is thrown horizontally, we only consider vertical motion to find the time it takes to hit the ground. The formula for time when initial vertical velocity is zero is derived from the equation of motion: \[ y = \frac{1}{2} g t^2 \]where \( y = 12.0 \) m is the height of the throw and \( g = 9.81 \) m/s² is the acceleration due to gravity. Solving for \( t \): \[ t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 12.0}{9.81}} \approx 1.56 \text{ s} \]

Step 3: Calculate the Horizontal Motion

The horizontal distance \( x \) that the ball travels is given by the equation \[ x = v_x \cdot t \]where \( v_x = 8.50 \) m/s is the horizontal velocity of the ball. Substituting the time \( t = 1.56 \text{ s} \) we found:\[ x = 8.50 \times 1.56 \approx 13.26 \text{ m} \]

Step 4: Determine Speed of the Dog

The dog must cover the same horizontal distance as the ball. His speed \( v_d \) must be equal to the horizontal velocity of the ball:\[ v_d = \frac{x}{t} = 8.50 \text{ m/s} \]

Step 5: Calculate the Distance from the Tree

The distance from the tree is the horizontal distance \( x \) where the ball lands, which we found to be approximately 13.26 m.

Key Concepts

Horizontal and Vertical Motion

When we talk about projectile motion, it's essential to break it into two components: horizontal and vertical motion. These two motions are independent of each other.
The horizontal motion of a projectile, such as the ball in our exercise, happens at a constant velocity because there are no forces acting on it horizontally (ignoring air resistance). In our case, the initial horizontal velocity is given as 8.50 m/s.
Vertical motion, on the other hand, is affected by gravity. When the ball is thrown horizontally from the tree, it immediately begins to fall due to gravity. At the start, there is no vertical velocity component; it is only gravity pulling it downward, causing it to accelerate as it falls. As a result, the ball's velocity increases as it descends.
This separation of components helps us understand how a projectile moves:

• The horizontal motion helps us determine the distance traveled.
• The vertical motion helps us calculate how long it will take to hit the ground.

Acceleration due to Gravity

Gravity is a force that acts on all objects with mass, pulling them towards the center of the Earth. The standard acceleration due to gravity is symbolized as \( g \) and is approximately 9.81 m/s². This is crucial in calculating the time it takes for a projectile to hit the ground.
In our problem, the ball is thrown from a height of 12 m. Since it is thrown horizontally, we start with zero initial vertical velocity. Using the equation of motion \( y = \frac{1}{2} g t^2 \), we can solve for the time \( t \) it takes the ball to reach the ground.
Plugging in the provided values, we find the time to be approximately 1.56 seconds. This means that from the moment the ball is thrown, it will take about 1.56 seconds for it to fall and reach the dog's level, illustrating how gravity affects all objects, causing them to accelerate downwards uniformly.

Kinematics Equations

Kinematics is the branch of physics that deals with motion without considering the forces that cause it. The equations of motion are handy tools that help us calculate various parameters such as displacement, velocity, acceleration, and time.
In solving projectile problems like ours, we often use several key kinematic equations. First, we use \( y = \frac{1}{2} g t^2 \) to find the time it takes for the projectile to fall to its target. Once we know the time, we can calculate how far it travels horizontally using \( x = v_x \cdot t \).
In our step-by-step solution, after determining the time (1.56 seconds), we used it to find the horizontal distance with the given constant horizontal speed (8.50 m/s). The final equation used illustrates how the kinematics equations provide all the necessary relations to fully describe the motion of projectiles in two dimensions.