Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

A ball is thrown upward from the surface of a planet where the acceleration of gravity is \(k\) (a negative constant) feet per second per second. If the initial velocity is \(v_{0}\), show that the maximum height is \(-v_{0}^{2} / 2 k\).

Step-by-Step Solution

Answer

The maximum height is \(-v_0^2 / 2k\).

Step 1: Identify the Equation of Motion

The motion of a ball thrown vertically can be described by the kinematic equation: \[ h(t) = v_0 t + \frac{1}{2} k t^2 \] where \( h(t) \) is the height at time \( t \), \( v_0 \) is the initial velocity, and \( k \) is the acceleration due to gravity.

Step 2: Compute the Velocity Function

The velocity of the ball is the derivative of the height function: \[ v(t) = \frac{d}{dt}(v_0 t + \frac{1}{2} k t^2) = v_0 + kt \] The ball reaches its maximum height when the velocity is zero.

Step 3: Set the Velocity to Zero to Find t

Set the velocity function to zero to find the time at which the ball reaches maximum height: \[ v_0 + kt = 0 \] Solving for \( t \), we get: \[ t = -\frac{v_0}{k} \]

Step 4: Determine the Maximum Height

Substitute \( t = -\frac{v_0}{k} \) into the height equation to find the maximum height: \[ h(-\frac{v_0}{k}) = v_0\left(-\frac{v_0}{k}\right) + \frac{1}{2} k \left(-\frac{v_0}{k}\right)^2\]Simplify this expression to find:\[ h_{max} = -\frac{v_0^2}{2k}\]

Step 5: Verify the Maximum Height Expression

The simplified expression \( h_{max} = -\frac{v_0^2}{2k} \) is verified to be correct, confirming the maximum height reached by the ball given the initial conditions.

Key Concepts

Kinematic Equations

In physics, kinematic equations describe the motion of objects without considering the forces that cause the motion. These equations help us predict various parameters like displacement, velocity, and acceleration. For projectile motion, a common kinematic equation is \[ h(t) = v_0 t + \frac{1}{2} k t^2 \] where:

• h(t) is the height or position as a function of time.
• v_0 represents the initial velocity.
• k is the acceleration due to gravity.

The presence of the constant term \(k\) affects how an object's velocity changes over time. This equation reflects how the initial speed and gravity's pull impact the height of an object thrown vertically. By utilizing this equation, it's possible to calculate the exact position of the object at any given time.

Derivative of Motion

To understand when an object reaches its maximum height, we consider its derivative of motion, which provides the velocity function. The derivative of the kinematic equation for height, \[ v(t) = \frac{d}{dt}(v_0 t + \frac{1}{2} k t^2) = v_0 + kt \] tells us the velocity at any moment in time. Here's the breakdown:

• The derivative of \(v_0 t\) is simply \(v_0\), allowing this term to remain constant.
• The derivative of \(\frac{1}{2} k t^2\) becomes \(kt\), showing how gravity influences velocity.

When the velocity function equals zero, the object reaches its peak height. Thus, by setting \(v(t) = 0\), and solving for \(t\), you find the time at maximum height. This is because an object at its maximum height in projectile motion has no vertical velocity in that instant.

Gravity on Other Planets

Gravity is a universal constant that varies from planet to planet, affecting how objects move and behave. Its strength is denoted in kinematic equations as \(k\), the acceleration due to gravity. For Earth, \(k\) typically is \(-9.8\) m/s², but on other planets, it differs significantly. Understanding how gravity impacts projectile motion involves:

• Recognizing that greater values of \(|k|\) lead to quicker deceleration of upward motion.
• Smaller \(|k|\) values mean a longer time for objects to reach their peak and a higher maximum height.
• The specific value of \(k\) depends on the mass and size of the planet.

By substituting different \(k\) values into our kinematic equations, we can predict and understand how objects will behave in non-Earth environments. This is crucial for space explorations and understanding celestial mechanics.

Initial Velocity

The initial velocity, often symbolized as \(v_0\), is the speed at which an object begins its journey in motion. It's a critical factor in calculating the maximum height of a projectile. Here is why:

• A higher \(v_0\) means the object will travel higher before gravity slows it to a stop.
• Determines both the ascent duration and speed for achieving peak height.
• Appears as a linear term in kinematic equations, directly affecting the overall motion.

In our problem, \(v_0\) is crucial because the maximum height formula \(-\frac{v_0^2}{2k}\) showcases how initial speed has a squared relationship with height. Double the initial velocity means a fourfold increase in maximum height, provided gravity remains constant. Understanding this relationship helps in predicting how far and high an object will go after being projected.