When a ball is thrown upwards, its motion is affected by gravity, which acts downwards. As the ball moves upwards, its speed decreases due to the opposing force of gravity. The ball reaches its maximum height when its instantaneous velocity becomes zero before it starts falling back down.

Understanding the Physics of Upward Motion

This problem involves understanding motion under constant acceleration, which is the acceleration due to gravity (\(g\)). For upward motion, we consider the acceleration due to gravity as negative because it opposes the direction of initial velocity.

Given Information

• Initial velocity of the ball, \(u = 2 \, \text{m/s}\)
• Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)

Goal

Find the maximum height (\(h_{max}\)) the ball reaches.

Applying Kinematics Equations

At the maximum height, the ball momentarily stops before changing direction. Therefore, the final velocity (\(v\)) at the maximum height is \(0 \, \text{m/s}\). We can use a standard kinematics equation that relates initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and displacement (\(s\)). The acceleration in this case is due to gravity, acting downwards, so \(a = -g = -10 \, \text{m/s}^2\). The displacement (\(s\)) will be the maximum height (\(h_{max}\)).

The relevant equation is:

\(v^2 = u^2 + 2as\)

Step-by-Step Calculation of Maximum Height

Substitute the known values into the equation:

\(0^2 = (2 \, \text{m/s})^2 + 2(-10 \, \text{m/s}^2)h_{max}\)

Simplify the equation:

\(0 = 4 \, \text{m}^2/\text{s}^2 - 20 \, \text{m/s}^2 \cdot h_{max}\)

Rearrange the equation to solve for \(h_{max}\):

\(20 \, \text{m/s}^2 \cdot h_{max} = 4 \, \text{m}^2/\text{s}^2\)

Divide both sides by \(20 \, \text{m/s}^2\) to find \(h_{max}\):

\(h_{max} = \frac{4 \, \text{m}^2/\text{s}^2}{20 \, \text{m/s}^2}\)

\(h_{max} = \frac{4}{20} \, \text{m}\)

\(h_{max} = \frac{1}{5} \, \text{m}\)

\(h_{max} = 0.2 \, \text{m}\)

Result

The maximum height the ball will reach is \(0.20 \, \text{m}\).

Revision Table: Maximum Height Calculation

Additional Information: Projectile Motion Concepts

Understanding projectile motion is key to solving problems like this. While this specific problem is one-dimensional vertical motion, it's a fundamental part of understanding projectile motion where objects move in two dimensions under gravity.

• Vertical Motion: Governed by gravity, causing constant downward acceleration (\(-g\)).
• Horizontal Motion (if any): In the absence of air resistance, horizontal velocity remains constant.
• Time of Flight: The total time the object spends in the air.
• Range: The horizontal distance covered by the projectile.

The equations of kinematics used here are derived from the definitions of velocity and acceleration assuming constant acceleration. For upward motion problems, always remember to use \(a = -g\) and \(v=0\) at the highest point.