Derive The Equation For Projectile Motion Under Gravity

Learn how to derive the key equations for projectile motion under gravity, including horizontal and vertical components, with step-by-step explanations and examples.

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Understanding Projectile Motion Basics

Projectile motion describes the path of an object launched into the air under the influence of gravity alone, with no air resistance. The motion separates into independent horizontal (x) and vertical (y) components. Horizontally, velocity remains constant (v_x = v_0 cos θ, where v_0 is initial speed and θ is launch angle). Vertically, acceleration is -g (g ≈ 9.8 m/s² downward), starting with v_y0 = v_0 sin θ. Deriving the equations involves applying kinematic formulas to each axis.

Deriving Horizontal Motion Equation

For the x-direction, acceleration a_x = 0, so the position equation simplifies to x = v_0 cos θ * t. This linear relationship shows the projectile travels a constant horizontal distance per unit time, forming the basis for range calculations.

Deriving Vertical Motion Equation

In the y-direction, use the kinematic equation y = v_y0 t + (1/2) a_y t², substituting v_y0 = v_0 sin θ and a_y = -g: y = (v_0 sin θ) t - (1/2) g t². To find the trajectory, eliminate t by solving x / (v_0 cos θ) = t and substituting into y, yielding y = x tan θ - (g x²) / (2 v_0² cos² θ). This parabolic equation describes the projectile's path.

Practical Applications and Importance

These derivations are essential in physics for predicting trajectories in sports (e.g., basketball shots), engineering (e.g., artillery), and space exploration. They enable calculations of maximum height (h = (v_0 sin θ)² / (2g)) and range (R = (v_0² sin 2θ) / g), optimizing designs and strategies in real-world scenarios.

Frequently Asked Questions

What assumptions are made in deriving projectile motion equations?
How do you find the time of flight for a projectile?
Why is the trajectory a parabola under gravity?
Does the derivation change if launched from a height?