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derive-maximum-height-formula

🚀 Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In two-dimensional motion, we analyze the motion of an object in a plane, which can be broken down into horizontal and vertical components. One important aspect of projectile motion is determining the maximum height reached by an object thrown upwards at an angle. The maximum height formula helps us calculate how high the object will go before it starts descending.

Theory Explanation

Understanding Projectile Motion

Projectile motion can be analyzed by separating the motion into horizontal and vertical components. The vertical motion is influenced by gravity, while the horizontal motion remains constant if air resistance is neglected.

Deriving the Maximum Height Formula

To derive the maximum height formula, we start with the vertical component of the initial velocity, which is given by \( v_{y0} = v_0 \sin(\theta) \). At the maximum height, the final vertical velocity \( v_y \) is 0. Using the kinematic equation \( v_y^2 = v_{y0}^2 - 2g h \), we can set \( v_y = 0 \) and solve for \( h \): \( 0 = (v_0 \sin(\theta))^2 - 2gh \). Rearranging gives us \( h = \frac{(v_0 \sin(\theta))^2}{2g} \). This is the maximum height formula.

\[ h = \frac{(v_0 \sin(\theta))^2}{2g} \]
Understanding Variables

In the formula, \( h \) is the maximum height, \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Understanding these variables is crucial for applying the formula correctly.

Key Points

  • 🎯 Kinematics focuses on the motion of objects without considering the forces involved.
  • 🎯 Projectile motion can be analyzed in two dimensions: horizontal and vertical.
  • 🎯 The maximum height formula is derived from the kinematic equations and is essential for solving projectile motion problems.

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Examples:💡

A ball is thrown upwards with an initial velocity of 20 m/s at an angle of 30 degrees. Calculate the maximum height reached by the ball.

Solution:

Step 1: Identify the initial velocity and angle: \( v_0 = 20 \, \text{m/s} \), \( \theta = 30^\circ \).

Step 2: Calculate the vertical component of the initial velocity: \( v_{y0} = v_0 \sin(\theta) = 20 \sin(30^\circ) = 20 \times 0.5 = 10 \, \text{m/s} \).

\[ v_{y0} = v_0 \sin(\theta) = 20 \times 0.5 = 10 \, \text{m/s}. \]

Step 3: Use the maximum height formula: \( h = \frac{(v_{y0})^2}{2g} = \frac{(10)^2}{2 \times 9.81} = \frac{100}{19.62} \approx 5.1 \, \text{m} \).

\[ h = \frac{(10)^2}{2 \times 9.81} \approx 5.1 \, \text{m}. \]

Common Mistakes

  • Mistake: Confusing the vertical and horizontal components of motion.

    Correction: Always separate the motion into vertical and horizontal components and apply the appropriate equations for each.

  • Mistake: Forgetting to convert angles to radians when using trigonometric functions in calculations.

    Correction: Ensure that angles are in the correct unit (degrees or radians) as required by the calculator or context.

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