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Projectile Motion overview

Topic 2 of 4

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 ax = 0 and horizontal velocity stays constant. Vertically, the only force is weight, so ay = -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 ux = u cos θ and uy = u sin θ. For a launch at position (x0, y0), the constant-acceleration equations give:

Horizontal

x = x0 + uxt
vx = ux

Uniform velocity because ax = 0.

Vertical

y = y0 + uyt - ½gt2
vy = uy - gt

Uniform acceleration -g with upward positive.

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, vy = 0. If ux 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.

Worked horizontal launch

Find time vertically, then range horizontally

A ball leaves a horizontal surface 1.25 m above the ground at 4.0 m/s. Use g = 10 m/s2. Put x0 = 0 and ground level at y = 0. Its initial components are ux = 4.0 m/s and uy = 0.

At landing, the vertical position is zero:

0 = 1.25 - ½(10)t2
t2 = 0.25 s2
t = +0.50 s

Select the positive time for landing after launch. The horizontal range is:

x = 4.0 × 0.50 = 2.0 m

At that same time, vx = 4.0 m/s and vy = 0 - 10(0.50) = -5.0 m/s. Combine perpendicular components to obtain the speed:

v = √(vx2 + vy2)
= √(4.02 + 5.02) = √41 m/s

The working values are speed 6.40 m/s and angle tan-1(5.0/4.0) = 51.3 degrees below horizontal. The signs of the components place the velocity down and right. To match the two-significant-figure inputs, report about 6.4 m/s and 51 degrees.

Different horizontal motion, matching vertical motion

Both balls start at height 1.25 m, with zero initial vertical velocity, at the same time. The model uses g = 10 m/s2 downward and negligible air resistance.

A horizontal launch and a matched vertical drop land togetherA calibrated position graph has equal horizontal and vertical scales, 120 drawing units per metre. The blue trajectory follows x equals four t and y equals 1.25 minus five t squared, for zero to 0.50 seconds. Its six filled dots are 0.10 seconds apart. Hollow brown dots at x zero show the vertically dropped ball at the same times and heights. The shared initial marker represents coincident initial coordinates in this comparison, not colliding balls in one experiment. Two horizontal guides connect matching heights at 0.20 and 0.40 seconds. The blue arrow at release is the four-metres-per-second horizontal initial velocity. A separate purple weight arrow points down on the launched ball during flight. Both reach ground level at 0.50 seconds; only the launched ball travels two metres horizontally. Arrow lengths do not share the position scale or compare velocity with force.00.511.5200.250.50.7511.25Height y / mHorizontal position x / mAt releaseu = 4.0 m/s rightWeight

Filled dots: launched ball. Hollow dots: dropped ball. Consecutive dots are 0.10 s apart; horizontal guides join two matching times. Both land at 0.50 s. The launched ball's range is 2.0 m.

The position axes have equal metre scales. The blue arrow shows initial velocity; the purple arrow is a force during flight. Neither arrow length is a position measurement.

Filled and hollow dots compare the two model motions at equal 0.1 s intervals. Matching vertical initial conditions give matching heights at each time, despite different horizontal motion.

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.

Optional check An ideal projectile has constant horizontal velocity 12 m/s right. At its highest point, take right and up as positive and g = 10 m/s^2. Which velocity and acceleration components are correct?
An ideal projectile has constant horizontal velocity 12 m/s right. At its highest point, take right and up as positive and g = 10 m/s^2. Which velocity and acceleration components are correct?

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.