Motion in a Plane · Lesson 8 of 13

Projectile Motion and the Parabolic Path

“Build the ideal projectile model from independent horizontal and vertical motion and derive its trajectory.”

Learning Objectives

• State the assumptions of the ideal projectile model. • Resolve launch velocity into horizontal and vertical components. • Calculate position, velocity, speed and direction at any time. • Derive the parabolic trajectory and explain motion at the highest point.

What changes after a ball leaves the hand?

A projectile is an object moving freely after launch, with gravity as the only significant force in the ideal model. The throwing hand no longer pushes it. Take a short enough flight near Earth’s surface that gravity is uniform and vertically downward, and neglect air resistance, wind effects and Earth’s rotation. These assumptions are useful for learning the motion; they can be poor approximations for a light object in strong drag or for a very long flight.

Choose x horizontal and y upward. The acceleration vector is −g j-hat: its horizontal component is zero and vertical component is −g. A zero horizontal acceleration means horizontal velocity remains constant. Vertical velocity changes exactly as it does in one-dimensional free fall. These two descriptions happen simultaneously and share one time variable.

Resolve the launch velocityLaTeX

u_{x} = u cos θ, u_{y} = u sin θ, a_{x} = 0, a_{y} = - g

θ is the launch angle above horizontal; u is launch speed. Keep signs if launching downward or leftward.

Position and velocity as functions of time

General launch pointLaTeX

x = x_{0} + u cos θ t, y = y_{0} + u sin θ t - \frac{1}{2} g t^{2}

Time zero is launch. Include y₀ when measuring height from the ground rather than from the launch point.

Velocity during flightLaTeX

v_{x} = u cos θ, v_{y} = u sin θ - g t, v = v_{x}^{2} + v_{y}^{2}

The direction is determined by the signed components, with tanφ = v_y/v_x and the correct quadrant.

At the highest point, v_y = 0. For a launch with a nonzero horizontal component, the total velocity remains horizontal and nonzero there. Acceleration is still vertically downward. This differs from a purely vertical throw, whose total velocity is zero at the top. During ascent, v_y is positive; during descent it is negative. The horizontal component remains unchanged throughout both phases.

Finding an intermediate point

Problem
A ball is launched with velocity 12 i-hat + 16 j-hat m/s from the origin. Take g = 10 m/s². Find position and velocity after 1 s.

1.Horizontal position is x = 12 × 1 = 12 m. Vertical position is y = 16 × 1 − ½ × 10 × 1² = 11 m.
2.Velocity components are v_x = 12 m/s and v_y = 16 − 10 = 6 m/s.
3.Speed is √(144 + 36) = 6√5 = 13.42 m/s. Direction is arctan(6/12) = 26.57° above horizontal.
4.The velocity direction is already shallower than the initial direction, although the horizontal component did not change.

Derive the equation of the path

Take x₀ = y₀ = 0 and assume u cosθ ≠ 0. The horizontal equation gives t = x/(u cosθ). Substitute this into the vertical equation. The first term becomes x tanθ and the second is proportional to x². A quadratic relation between y and x describes a parabola.

Trajectory for launch from the originLaTeX

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

This is a downward-opening parabola for g > 0 and a nonzero horizontal launch component. For a launch point (x₀,y₀), replace x by x − x₀ and y by y − y₀.

A projectile trajectory— Launch components are 12 m/s horizontally and 16 m/s upward, with g = 10 m/s². The peak is at x = 19.2 m, y = 12.8 m; the ball returns to y = 0 at x = 38.4 m.

The graph above is a spatial trajectory: both axes are positions. Unlike an x–t graph, its shape really does show the path through space. A tangent to it gives the local velocity direction. Its slope dy/dx equals v_y/v_x wherever v_x is nonzero. At the peak the tangent is horizontal, while acceleration remains downward.

Initial conditions select the trajectory

The same acceleration −g j-hat can produce several paths. A released object falls vertically; a vertically thrown object moves on one vertical line; a ball with a horizontal launch component follows a parabola. The acceleration by itself does not select the path. Initial velocity and initial position are essential. A purely vertical launch cannot be obtained by dividing by u cosθ = 0 in the trajectory formula; use its time equations directly.

Independent fall does not mean independent clocks

A ball dropped from rest and another launched horizontally from the same height have the same vertical initial velocity and acceleration. In the ideal model they reach the same lower horizontal plane after the same time. The horizontally launched ball travels farther sideways but does not “stay up” because of its horizontal motion. If a projectile has an upward initial component, its vertical initial condition is different and its flight time changes.

Quick check

At the highest point of an obliquely launched ideal projectile, which statement is correct?

Key Takeaways

• Ideal projectile acceleration is constant and vertically downward. • Horizontal velocity is constant; vertical velocity changes with time. • Use the same time in both coordinates. • The trajectory is parabolic when the horizontal launch component is nonzero. • At the top, zero vertical velocity does not mean zero total velocity or zero acceleration.

Previous · Lesson 7

Relative Motion in Two Dimensions

Next · Lesson 9

Projectile Height, Flight Time and Range