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 8.50 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?

Short answer

(a) Dog's speed: 8.53 m/s. (b) Distance: 13.3 meters.

Step-by-step solution

Identify Given Information

We know the initial height of the ball is 12.0 m, and it is thrown horizontally at a speed of 8.50 m/s. The initial vertical velocity of the ball is 0 since it is thrown horizontally.

Determine Time of Flight

To find out how long the ball is in the air, we use the formula for vertical motion due to gravity: \[ h = \frac{1}{2} g t^2 \]where \( h = 12.0 \) m (height), and \( g = 9.81 \) m/s² (acceleration due to gravity). Solving for time \( t \):\[ 12 = \frac{1}{2} \times 9.81 \times t^2 \]\[ t^2 = \frac{24}{9.81} \]\[ t = \sqrt{\frac{24}{9.81}} \approx 1.56 \text{ seconds} \]

Calculate Horizontal Distance

The horizontal distance the ball travels, which is also the distance the dog must run, can be found using the formula:\[ \text{Distance} = \text{velocity} \times \text{time} \]\[ \text{Distance} = 8.50 \times 1.56 \approx 13.3 \text{ meters} \]

Determine Dog's Required Speed

To catch the ball, the dog must match the horizontal distance in the same time. Therefore, the dog's speed \( v_d \) is:\[ v_d = \frac{\text{Distance}}{\text{time}} = \frac{13.3}{1.56} \approx 8.53 \text{ m/s} \]

Key concepts

Kinematics

Kinematics is the branch of physics that focuses on the motion of objects without considering the forces that cause this motion. A kinematic description relies on the concepts of displacement, velocity, and acceleration. In the given problem, we analyze the motion of a ball thrown horizontally from a height, aiming to understand both its path and the motion needed by the dog to catch it.

When dealing with kinematics, it is essential to separate vertical and horizontal motions, as they do not affect each other. The total motion is a combination of both with each having unique contributing factors. The horizontal movement relates to the speed at which the ball is thrown, while the vertical aspect is influenced by gravity. Together, these components help us predict where and when the ball will land.

Horizontal and Vertical Components

Projectile motion involves two key components: horizontal and vertical motion. These components operate independently due to the perpendicular nature of forces acting upon them.

• Horizontal component: In this exercise, the ball is thrown horizontally at 8.50 m/s. This constant velocity determines how far the ball will travel horizontally while it is in the air.
• Vertical component: The vertical motion is influenced by gravitational acceleration, which affects the time the ball remains airborne. Initially, the vertical velocity is 0 m/s, because the ball is thrown horizontally.

The independence of these two components means that horizontal velocity remains constant (as there is no horizontal acceleration in an ideal scenario), while vertical velocity changes due to gravity.

Acceleration due to Gravity

The acceleration due to gravity is a critical factor in projectile motion, playing a vital role in determining the time a projectile remains in the air. For objects close to the surface of the Earth, this acceleration is approximately constant at 9.81 m/s² downward.

In the vertical component of motion, gravity causes the ball's speed to increase downwards as it falls. The formula we use to calculate the time it takes for the ball to hit the ground is derived from the basic kinematic equation for free fall: \[ h = \frac{1}{2} g t^2 \]Solving this equation gives the time of flight, allowing us to further calculate how far horizontally the ball will travel before it hits the ground.

The precise calculation of how gravity affects vertical motion also helps to predict the ball's trajectory, ensuring accurate determination of landing points in projectile problems like the one presented.