3.1.3 - Projectile motion

The Projectile Model

A projectile is an object moving through the air after it has been launched, thrown, dropped or released, when the only force we include in the model is its weight. In OCR problems this usually means air resistance is negligible unless the question explicitly says otherwise.

In this model, gravity acts vertically downwards. There is no horizontal force, so there is no horizontal acceleration. That gives the central idea:

Independence means the horizontal velocity does not make the object fall faster or slower. If two balls leave the same height at the same time, one dropped vertically and one launched horizontally, they have the same vertical motion and land at the same time, provided air resistance is negligible.

[DIAGRAM: asset_name: 3.1.3 - Projectile motion - diagram 1; asset_slug: 3_01_3_projectile_motion__diagram_01; recommended_method: drawn_physics; caption: Equal-time positions show constant horizontal velocity and increasing downward speed.; description: Equal-time positions show independent horizontal and vertical motion: horizontal steps are equal while vertical drops increase.]

The most common trap is at the highest point of a flight. The vertical velocity is zero for an instant there, but the acceleration is still $9.81\,\text{m s}^{-2}$ downwards. The projectile is still a projectile; gravity has not switched off.

Resolving The Launch Velocity

Most projectile questions begin with an initial speed at an angle. Before using any motion equation, split the initial velocity into horizontal and vertical components.

If a projectile is launched with initial speed $u$ at an angle $\theta$ above the horizontal:

Here $u_x$ is the initial horizontal velocity and $u_y$ is the initial vertical velocity. This formula box assumes the angle is measured from the horizontal. If the angle is measured from the vertical, redraw the right-angled triangle before deciding which component uses sine or cosine.

Once the components are known, treat the two directions separately. If up is chosen as positive, then:

The symbol $x$ is horizontal displacement, $y$ is vertical displacement, $t$ is time, $v_x$ is horizontal velocity, and $v_y$ is vertical velocity. Use $g = 9.81\,\text{m s}^{-2}$ unless the question gives another value.

Worked example: a ball is kicked at $24.0\,\text{m s}^{-1}$ at $35.0^\circ$ above the horizontal. Find its initial horizontal and vertical velocity components.

The horizontal component is larger because the angle is closer to the horizontal than to the vertical.

Horizontal Launches

For a horizontal launch, the initial vertical velocity is zero:

The vertical motion then looks exactly like free fall from rest. The horizontal motion is constant-velocity motion. The time is usually found from the vertical information, then used in the horizontal equation.

Worked example: a small ball rolls horizontally off a bench of height $1.25\,\text{m}$ with a horizontal speed of $7.20\,\text{m s}^{-1}$. Calculate the horizontal distance from the bench where it lands. Ignore air resistance.

Choose downward displacement as positive for the vertical part:

Use the vertical equation first:

Now use horizontal constant velocity:

So the ball lands $3.64\,\text{m}$ horizontally from the bench.

Notice the logic. The horizontal speed did not affect the time of fall. It only affected how far the ball travelled horizontally during that time.

Angled Projectiles

For a projectile launched at an angle, the vertical velocity changes throughout the flight:

• it starts upward if $u_y$ is positive
• it becomes zero at maximum height
• it becomes downward as the projectile falls

The horizontal velocity stays constant throughout the flight, if air resistance is negligible.

At maximum height:

This is a useful condition, not a new force or a new equation. The acceleration is still downward.

Worked example: the ball from the previous component example is launched at $24.0\,\text{m s}^{-1}$ at $35.0^\circ$ above the horizontal. It lands at the same vertical height from which it was launched. Calculate the time to maximum height, the maximum height, and the horizontal range.

Keep the full calculator values of the launch components:

At maximum height $v_y=0$, so $0=u_y-gt$:

Using $v_y^2=u_y^2-2gy$, the maximum height is

Because landing is at the launch height, the total time is twice the unrounded time to the top:

The horizontal range is

Do not memorise a special range formula for OCR questions here. The safer method is to resolve, find time from the vertical motion, then use the horizontal motion.

Projectile paths, velocities and video data

A position at a known horizontal distance is found by using horizontal motion to determine the shared time, then calculating vertical displacement. This also handles targets above or below the launch height, where doubling the time to the top would be invalid.

Worked example: a ball is projected at $28.0\,\text{m s}^{-1}$ at $18.0^\circ$ above horizontal ground. A vertical line is $45.0\,\text{m}$ from the launch point. Calculate the height of the ball above the launch level when it reaches that line. Ignore air resistance.

Resolve first, retaining full calculator values:

The horizontal displacement determines the shared time:

Use this unrounded time in $y=u_yt-\frac12gt^2$:

The ball is about $0.615\,\mathrm{m}$ above the launch level at this horizontal position. Rounding the components or time early changes the small difference between the two vertical-displacement terms substantially.

Although OCR 3.1.3 has no named PAG, projectile motion can be assessed through practical and data-handling contexts. A sensible school investigation might use a ball launcher, metre rule, vertical screen, video analysis or light gates to record positions at known time intervals. The independent variable might be launch angle or initial speed; the dependent variable might be range or height at a fixed horizontal distance. Control variables could include launch height, launch position, ball type and surface level.

Uncertainty reasoning should match the apparatus. A ruler reading may have an uncertainty based on its scale resolution; a stopwatch measurement is often limited by reaction time; video analysis reduces reaction-time error but introduces scale calibration, frame-rate and parallax uncertainties. Repeating launches, using a fixed launcher, marking the same release point, calibrating the video scale, and measuring over larger distances can reduce percentage uncertainty.

To find the velocity at a later instant, use $v_x=u_x$, $v_y=u_y-gt$, then $v=\sqrt{v_x^2+v_y^2}$. Its angle to the horizontal satisfies $\tan\alpha=|v_y|/|v_x|$; state whether the direction is above or below horizontal. For $v_x=12.0\,\mathrm{m\,s^{-1}}$, $v_y=-5.0\,\mathrm{m\,s^{-1}}$, the speed is $13.0\,\mathrm{m\,s^{-1}}$ at $22.6^\circ$ below horizontal. A negative vertical component is a direction, not a negative speed.

In equal-time video frames, equal horizontal position changes indicate constant horizontal velocity. Increasing vertical changes indicate increasing vertical speed. Only equal increases in those changes establish constant vertical acceleration; increasing spacing alone does not establish a constant value.

[DIAGRAM: asset_name: 3.1.3 - Projectile motion — figure 2; asset_slug: 3_01_3_projectile_motion__diagram_02; recommended_method: drawn_physics; caption: Projectile velocity has perpendicular components while gravitational acceleration acts downward.; description: A projectile trajectory with launch velocity resolved into perpendicular horizontal and vertical components; gravitational acceleration acts downward.]