Parametric Equations for Projectile Motion: A Khan Academy Example

Setting Up Parametric Equations for Projectile Motion

• 🎯 The problem involves launching a water balloon with an initial velocity of 25 m/s at a 35° angle from a height of 1 meter.
• 🚀 To model the path, we break the initial velocity into its horizontal (x) and vertical (y) components using trigonometry.
• 💡 The horizontal component of velocity is calculated as 25 * cos(35°), and the vertical component as 25 * sin(35°), assuming no air resistance or friction.

Deriving the Position Functions

• 📈 The x-position as a function of time x(t) is simply the horizontal velocity multiplied by time: x(t) = (25 * cos(35°)) * t.
• 📉 The y-position as a function of time y(t) incorporates gravity and the initial height: y(t) = (25 * sin(35°)) * t - (1/2) * 9.8 * t^2 + 1.
• ⚠️ We assume x(0) = 0 and y(0) = 1 meter for the initial conditions.

Calculating Height at the Target Distance

• 🎯 To find the height when the balloon reaches the target 55 meters away, we first solve for the time t when x(t) = 55.
• ⏱️ This yields t = 55 / (25 * cos(35°)), which is approximately 2.686 seconds.
• 📊 Plugging this time back into the y(t) equation gives the height at that point: y(2.686) = (25 * sin(35°)) * 2.686 - 4.9 * (2.686)^2 + 1.
• ✅ The calculated height of the water balloon when it reaches the target distance is approximately 4.2 meters.