Projectile Motion

01

Weight and gravitational energy change

Weight is the gravitational force on a mass. A change in height changes the gravitational energy of the mass-Earth system; it does not change the body's mass.

Mass m is measured in kg, weight W in N, and gravitational field strength g in N/kg. Weight acts along the gravitational field. Here the field is downward and effectively uniform over the height range.

When gravity is the only significant force and mass is constant, Newton's second law gives ma = mg, so the downward acceleration has magnitude g. With upward positive, a_y = -g. The units N/kg and m/s² are equivalent, but force per unit mass and acceleration describe different quantities.

A support or scale reading is not always the weight. For a body with only upward support R and downward weight mg, Newton's second law gives R - mg = ma_y. The reading R equals mg only when vertical acceleration is zero. An unsupported body can fall freely with R = 0 while gravitational weight remains.

Derive the gravitational energy change

Choose upward as positive and let Δh = h_final - h_initial. The constant gravitational force is downward, opposite a positive height change. From force multiplied by displacement in its direction:

For the work of a static gravitational field, W_gravity = -ΔE_p. Combining the two statements gives:

This is the change in the mass-Earth system's gravitational potential energy. It does not assign an absolute zero at every location. Adding a common constant to the initial and final energies leaves the change unchanged.

Another way to see the result is to lift the mass slowly at constant speed with no other energy changes. The lifting force is mg, so the external agent's work is mgΔh. That positive transfer supplies the gravitational increase. If kinetic energy changes or energy dissipates, the agent's total work need not equal the gravitational increase alone.

Only the vertical displacement contributes to work by the vertical gravitational force. Do not put a sloping path length into mgΔh. Friction or a kinetic-energy change can alter the required external work without altering the gravitational change between the stated heights.

The formula assumes an effectively uniform field. It is suitable for small height changes near Earth's surface, not as one constant-g expression for all distances from Earth.

02

Perpendicular components, one elapsed time

Describe horizontal and vertical motion separately, then combine their positions or velocities at the same time. The two equations describe one moving body.

Start with the force model

During ideal projectile flight near Earth, the body has left its launcher or support. Assume a uniform downward gravitational field, negligible air resistance and no other significant force. There is no continuing forward launch force.

Choose right and up as positive. Weight has no horizontal component, so a_x = 0 and horizontal velocity stays constant. Vertically, the only force is weight, so a_y = -g. Both components use the same elapsed time t, measured from launch.

If the launch speed is u at an angle θ above the positive horizontal, its components are u_x = u cos θ and u_y = u sin θ. For a launch at position (x₀, y₀), the constant-acceleration equations give:

Position (x, y) traces the flight path. It is not a velocity-time graph. The body's velocity is tangent to the trajectory, while its weight acts downward. These arrows describe different quantities.

At the highest point of an upward launch, v_y = 0. If u_x is nonzero, the body is still moving horizontally. Its downward acceleration remains g: gravity does not switch off when the vertical velocity is momentarily zero.

Estimate the scale of a horizontal flight

Assume a roughly 0.2 kg ball leaves horizontally at about 4 m/s from about 1 m above the ground. Take uniform g ≈ 10 m/s² and negligible drag. The vertical drop gives t ≈ √(2h/g) ≈ 0.45 s: roughly half a second.

Horizontal travel is then roughly 4 × 0.45 ≈ 2 m, and the gravitational change is roughly -mgh ≈ -2 J. These estimates use the stated rough height, speed and mass; they give a scale against which to check a more precise calculation.

Compare with a vertical drop

A second ball released from rest at the same height at the same instant has the same initial vertical velocity, zero, and the same vertical acceleration, -g. It therefore reaches the same ground level in 0.50 s under this model.

The horizontal component does not reduce gravity or delay the fall. Matching masses alone is not the reason for equal landing times. A ball thrown upward has a different initial vertical velocity and does not share this result merely because it has the same mass.

Test the model with calibrated video

A position record can test whether horizontal velocity is approximately constant and vertical acceleration is approximately -g. For a suitable video:

1. Fix the view. Keep the camera stationary, facing perpendicular to the motion plane. Put a length scale in that same plane, and identify the true horizontal and vertical directions.
2. Keep one position reference. Track the same point, such as the ball's centre, in each frame. Record its coordinates from a fixed origin; use clear images to limit uncertainty from motion blur.
3. Use actual times. Record frame timestamps or the known interval between selected frames. A skipped frame is a larger interval, not another ordinary step.
4. Form interval velocities. Calculate Δx/Δt and Δy/Δt, and associate each average with its interval's midpoint time. For uniform acceleration this equals the instantaneous velocity at that midpoint.
5. Compare trends. Equal horizontal changes in equal times support constant horizontal velocity. Changes in vertical velocity divided by the actual time difference should be approximately -g under the ideal model.

The position-data method gives a worked example of midpoint velocities. Differencing uncertain positions can produce noisy velocities, and differencing those again can magnify the scatter in acceleration. Use suitable intervals and compare the overall trend rather than interpreting one fluctuation as a new force.

A scale outside the motion plane or an oblique camera view can distort distances. If each pixel is assigned too large a distance, inferred lengths and speeds are systematically too large. Treating two frame intervals as one makes the velocity calculation divide by too short a time. Repeated readings do not remove either wrong calibration or wrong timing.

Appreciable drag or wind, especially over a longer flight during which their effects accumulate, can undermine the ideal equations. Check whether calibrated observations support the force assumptions. A near-terminal fall is a different model from this gravity-only flight.

03

Launch and land at different heights

Set the vertical equation equal to the actual landing height. Returning to the launch height and reaching lower ground are different events.

Assume ideal flight: uniform downward g = 10 m/s², negligible air resistance and no force after launch except weight. Choose right and up as positive, with t = 0 at launch and ground level at y = 0.

The initial position is (0, 20 m), with u_x = 12 m/s and u_y = 9.0 m/s. These components give launch speed √(12² + 9²) = 15 m/s at tan^-1(9/12) = 36.9 degrees above horizontal.

As in the horizontal-launch example, horizontal velocity is constant and vertical acceleration is -g. With t in s, the numerical equations give positions in m and velocities in m/s:

1. Choose the future landing root

The ground is y = 0, so:

The roots are approximately +3.093 s and -1.293 s. Only the positive root describes the landing after the stated launch. The negative root belongs to a mathematical extension of the equation to an earlier time; it is not this flight's landing time.

The horizontal range is 12 × 3.093171... = 37.12 m as a working value. Keep the unrounded time while calculating later quantities.

2. Combine the landing velocity components

The negative vertical component means downward motion. Speed and direction follow from both components:

The landing direction is 61.3 degrees below horizontal, down and right. The result 21.93 m/s is the magnitude of the vertical component, not the full speed.

3. Distinguish the top and return to launch height

At the top, v_y = 0, so 9 - 10t = 0 gives t = 0.90 s. Its position is x = 12(0.90) = 10.8 m and y = 20 + 9(0.90) - 5(0.90)² = 24.05 m. Speed is still 12 m/s because the horizontal component remains.

On descent, the projectile returns to y = 20 m when 9t - 5t² = 0 gives the nonzero time 1.80 s. It is then at x = 21.6 m, still above the ground. The expression 2u_y/g gives this same-height return, not the time to the lower landing level.

4. Check the speed using energy

Let the ball's mass be 0.20 kg. With negligible drag, the ball-Earth system's decrease in gravitational potential energy becomes an increase in kinetic energy during flight.

Then v = √(2E_k/m) = √(2 × 62.5/0.20) = 25.0 m/s, agreeing with the component calculation. Energy finds the speed here; it has not by itself found the flight time or direction.

The extra digits in these model calculations help compare methods and locate graph features. To two significant figures, the landing results are 3.1 s, 37 m, 25 m/s and 61 degrees below horizontal. Retain the unrounded time for intermediate calculations, and report final results to precision justified by the supplied quantities.

Read the equal-time position data

Every 0.50 s adds 6.0 m horizontally, giving Δx/Δt = 12 m/s. Vertical changes are different: from 0 to 0.50 s, average vertical velocity is (23.25 - 20.00)/0.50 = 6.5 m/s, associated with midpoint time 0.25 s. From 0.50 to 1.00 s it is (24.00 - 23.25)/0.50 = 1.5 m/s, at midpoint time 0.75 s.

The velocity change divided by the midpoint-time difference is (1.5 - 6.5)/(0.75 - 0.25) = -10 m/s². This agrees exactly with the uniform-acceleration model used to calculate the table. Real video readings have uncertainty and must be assessed using the calibration and timing method; a generated table is not experimental verification.

04

Falling with air resistance

A falling body's weight can stay constant while its acceleration changes. As drag grows, the resultant force decreases; at terminal speed the body keeps moving while the forces balance.

Consider a body released from rest in still air. Use a uniform gravitational field, constant mass and negligible buoyancy. Its shape and the surrounding air conditions remain unchanged, and drag increases with its speed relative to the air.

For this fall, choose downward as positive. This differs from the upward-positive convention used for the projectile examples. Let D be the upward drag magnitude:

1. Just released: relative air speed is zero, so drag is zero in this model. The downward resultant is mg and downward acceleration is g.
2. Speed increasing: drag acts upward and grows. It is smaller than weight, so the body still accelerates downward, but its acceleration is less than g.
3. Terminal motion: drag balances weight. The resultant and acceleration are zero while the body continues downward at a nonzero constant speed.

The speed-time gradient is the downward acceleration during this downward motion. The curve becomes less steep as drag grows and tends to a horizontal line at terminal speed. The body does not stop there. The unchanged-condition model does not predict a speed overshoot followed by an upward acceleration.

The constant-acceleration equations cannot be used across the whole drag-dominated fall: a = g - D/m changes as D changes. Near terminal motion, acceleration is approximately zero even though gravitational field strength remains nonzero.

Follow the energy as well as the forces

Before terminal speed, decreasing gravitational potential energy supplies both increasing kinetic energy and increasing internal energy of the body and air. The kinetic-energy increase is smaller than the gravitational decrease when energy is dissipated.

At terminal speed, kinetic energy is constant, but the body continues descending. Gravitational potential energy continues to decrease, and energy continues transferring to internal stores. A zero resultant is not zero work by each individual force.

If the drag conditions change

Opening a parachute changes the relationship between drag and speed. Immediately after opening, the upward drag may exceed weight while the body is still moving down. The upward resultant slows its downward motion; it does not make velocity reverse instantly.

As speed decreases, drag falls towards a new balance at a lower terminal speed. With downward positive, the slowing stage has negative acceleration. This changed-condition example differs from the earlier approach to terminal speed for an unchanged body.