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 \(-v_{0}^{2} / 2 k\).

Short answer

The maximum height is \\(-\frac{v_{0}^2}{2k}\\)."

Step-by-step solution

Understand the Problem

We need to find the maximum height of a ball thrown upward from a planet with gravity acceleration \(k\) given the initial velocity \(v_{0}\). The gravity \(k\) is a negative constant, describing the downward force.

Determine the Ball's Motion Equation

The motion of the ball can be described by the equation \(h(t) = v_{0}t + \frac{1}{2}kt^2\), where \(h(t)\) is the height at time \(t\), \(v_{0}t\) represents the initial velocity impact, and \(\frac{1}{2}kt^2\) represents the gravity's effect.

Find the Velocity as a Function of Time

The velocity \(v(t)\) is the derivative of the height function with respect to time. Therefore, \(v(t) = \frac{dh}{dt} = v_{0} + kt\).

Find the Time at Maximum Height

The maximum height occurs when the velocity is zero. Set \(v(t) = 0\), so \(v_{0} + kt = 0\). Solving for \(t\), we find \(t_{max} = -\frac{v_{0}}{k}\).

Calculate Maximum Height

Substitute \(t_{max} = -\frac{v_{0}}{k}\) back into the height equation \(h(t) = v_{0}t + \frac{1}{2}kt^2\) to find the maximum height: \(h_{max} = v_{0}(-\frac{v_{0}}{k}) + \frac{1}{2}k(-\frac{v_{0}}{k})^2\).

Simplify the Maximum Height Expression

Simplify the expression for \(h_{max}\): \(-\frac{v_{0}^2}{k} + \frac{1}{2}k\Big(\frac{v_{0}^2}{k^2}\Big) = -\frac{v_{0}^2}{k} + \frac{v_{0}^2}{2k} = -\frac{2v_{0}^2}{2k} + \frac{v_{0}^2}{2k} = -\frac{v_{0}^2}{2k}\)."

Verify the Final Expression

We've derived that the maximum height \(h_{max}\) is \(-\frac{v_{0}^2}{2k}\), which matches the required solution.

Key concepts

Projectile Motion

Projectile motion refers to the path an object follows when it is thrown or propelled into the air. When we think about projectile motion, we often imagine a familiar arch-like curve. This is because projectile motion has two components: horizontal and vertical. In the horizontal direction, any motion is typically constant in speed unless another force acts on it. However, in vertical motion, gravity plays a crucial role.

In this specific problem where a ball is thrown upward, the initial velocity is strong enough to work against the pull of gravity, allowing the ball to ascend before gravity eventually overcomes this initial push, causing the ball to fall back down. The ball reaches its peak when its upward velocity reduces to zero, marking the maximum height. This instant is crucial as it helps us determine the maximum height using kinematic equations.

Gravity's Effect on Projectile

Gravity is a fundamental force pulling objects towards the center of a celestial body, like Earth. For this exercise, the gravitational pull is represented by a constant, denoted as \(k\) and is always negative, indicating a downward direction.

Gravity directly affects the projectile by continually decreasing its vertical velocity as it moves upward. This deceleration continues until gravity entirely negates the object's initial upward velocity, halting its rise.

• During the ascent, the ball's velocity gradually slows.
• At the peak, the velocity is briefly zero before gravity pulls it back down.
• On descent, the velocity increases in the opposite direction.

This negative gravitational acceleration is crucial when solving projectile motion problems using kinematic equations, as it determines how the object's velocity changes over time.

Kinematic Equations

Kinematic equations are mathematical formulas used to describe the motion of objects without considering the forces that cause this motion directly. They help us predict quantities such as displacement, velocity, and time in projectile motion.

In this problem, the height of the projectile ball is given by the equation \(h(t) = v_{0}t + \frac{1}{2}kt^2\). This formula includes:

• \(v_{0}t\), representing the initial velocity's contribution to height.
• \(\frac{1}{2}kt^2\), representing the effect of gravity over time.

The velocity equation \(v(t) = v_{0} + kt\) shows how velocity changes over time.
To find the maximum height, we set velocity \(v(t)\) to zero to find when the ball stops rising, solving for time gives us \(t_{max} = -\frac{v_{0}}{k}\).

Plugging \(t_{max}\) back into the height equation, and simplifying, gives us the formula for the maximum height \(h_{max} = -\frac{v_{0}^2}{2k}\). This illustrates the power of kinematic equations in solving real-world physics problems.