Problem 51 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A ball is thrown horizontally from the top of a cliff with initial speed \(v_{0}\) (at \(t=0 )\) . At any moment, its direction of motion makes an angle \(\theta\) to the horizontal (Fig. 47\()\) . Derive a formula for \(\theta\) as a function of time, \(t,\) as the ball follows a projectile's path.

Step-by-Step Solution

Answer

The formula is \( \theta = \arctan\left(\frac{gt}{v_0}\right) \).

Step 1: Understand the Components of Motion

A ball thrown horizontally means that the initial vertical velocity component is zero. Only after it is released does gravity act on it, giving it a velocity component downward.

Step 2: Define the Velocity Components

The horizontal component of velocity remains constant at \( v_0 \), while the vertical component of velocity at any time \( t \) is given by \( v_y = gt \), where \( g \) is the acceleration due to gravity.

Step 3: Calculate the Tangent of the Angle

The tangent of the angle \( \theta \) formed by the velocity with the horizontal axis is given by the ratio of the vertical component to the horizontal component: \( \tan(\theta) = \frac{v_y}{v_0} = \frac{gt}{v_0} \).

Step 4: Derive the Formula for \( \theta \)

By taking the arctangent on both sides, we get \( \theta = \arctan\left(\frac{gt}{v_0}\right) \). This gives the angle \( \theta \) as a function of time \( t \).

Key Concepts

Understanding Horizontal Velocity

In projectile motion, particularly when an object is thrown horizontally like the ball in our exercise, understanding horizontal velocity is key. Horizontal velocity (\( v_0 \)) refers to the constant component of velocity along the horizontal axis.
This remains constant throughout the motion, assuming no air resistance.The key points about horizontal velocity in our scenario are:

• Initial horizontal velocity is represented by \( v_0 \), the speed at which the ball is thrown from the cliff.
• Gravity does not affect horizontal velocity, meaning it doesn't change due to gravity.
• Horizontal velocity impacts how far the projectile will travel horizontally.

Remember, because the horizontal speed stays constant in ideal conditions (ignoring air resistance), the horizontal distance covered over time is straightforward to calculate using simple formulas.

Exploring Vertical Velocity

Vertical velocity is another critical component of projectile motion. Unlike horizontal velocity, vertical velocity changes over time due to the influence of gravity. When a projectile is thrown horizontally, its initial vertical velocity is zero because the motion starts with a horizontal push only.
Here's how vertical velocity behaves:

• Vertical velocity starts at zero but accelerates downward due to gravity.
• At any point in time \( t \), the vertical velocity \( v_y \) is given by \( v_y = gt \), where \( g \) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\) on Earth).
• Vertical velocity is responsible for the object's increase in speed downward as it moves along its trajectory.

Knowing how vertical velocity acts helps us understand how an object gains speed in the downward direction while moving along a projectile's path.

The Angle of Projection

The angle of projection, often represented by \( \theta \), is the angle between the velocity vector of the projectile and the horizontal line. This angle gives us insights into the direction of the projectile at any point in time. For a ball thrown horizontally, this angle initially starts at 0° and grows over time as vertical velocity increases.Key information about the angle of projection includes:

• The angle \( \theta \) is determined by the tangent of the vertical to horizontal velocity components: \( \tan(\theta) = \frac{v_y}{v_0} \).
• Given \( v_y = gt \) and constant \( v_0 \), the angle can be calculated as \( \theta = \arctan\left(\frac{gt}{v_0}\right) \).
• This angle is crucial for understanding how the projectile's path becomes steeper over time.

Grasping how the angle of projection changes helps explain the increasingly steeper trajectory of the projectile the longer it remains in motion.