SACE Physics Mastery Pack

1. Build the vector model before choosing equations

A projectile is an object whose motion after launch is governed chiefly by gravity. In the ideal near-Earth model, gravitational acceleration is constant and downward and air resistance is negligible. Choose a horizontal x-axis and a vertical y-axis, state the origin and declare positive directions. Taking right and up as positive gives a_x = 0 and a_y = −g. These statements describe acceleration, not velocity: a projectile can move upwards while accelerating downwards.

Resolve a launch speed u at angle θ above horizontal into u_x = u cos θ and u_y = u sin θ. For a launch at 20.0 m s⁻¹ and 30.0°, the components are 17.3 m s⁻¹ and 10.0 m s⁻¹. Their magnitudes do not add to 20.0 because they are perpendicular; √(17.3² + 10.0²) recovers the speed. A component can be negative when the corresponding motion opposes the chosen positive axis.

The horizontal and vertical motions share the same elapsed time. They are solved independently because acceleration in one direction does not directly change the perpendicular velocity component in this model. The equations are x = u_xt, y = u_yt − ½gt², v_x = u_x and v_y = u_y − gt. The vertical coordinate y is a displacement from the selected origin, not necessarily height above the ground.

Before substituting, make a small diagram listing the launch and landing coordinates. If the launch is from a platform 12.0 m above the landing surface and the origin is at launch, the landing condition is y = −12.0 m. Writing y = +12.0 would model a different event and can produce an apparently plausible but physically incorrect time.

2. Solve flight time using the landing condition

Flight time comes from the event that ends the flight. The shortcut t = 2u_y/g applies only when launch and landing are at the same height and drag is negligible. For a horizontal launch, u_y = 0; the object still falls because a_y = −g. For a launch above or below the eventual landing level, substitute the actual vertical displacement into y = u_yt − ½gt² and solve the resulting quadratic.

Consider the 20.0 m s⁻¹, 30.0° launch from a platform 12.0 m above the landing level. With g = 9.80 m s⁻², the landing equation is −12.0 = 10.0t − 4.90t², or 4.90t² − 10.0t − 12.0 = 0. The roots are [10.0 ± √(100 + 235.2)]/9.80. They give t = 2.889 s and a negative time of approximately −0.848 s. The positive root is the landing after launch.

The negative root is not evidence that the calculator failed. It lies on the mathematical extension of the trajectory before the chosen t = 0 and does not describe this flight. Reject it by referring to the physical domain t ≥ 0. If two positive times arise for a specified intermediate height, one can represent ascent and the other descent; the question must determine which event is relevant.

Use the accepted time in the horizontal equation: x = 17.3205 × 2.8886 = 50.0 m to three significant figures. A reasonableness check supports the result: flight to the launch height would take about 2.04 s, so landing 12.0 m lower must occur later and farther away. This comparison can reveal a sign error before it reaches the final answer.

3. Distinguish the apex from the end of motion

At the highest point, only the vertical component of velocity is zero. The horizontal component remains u_x in the ideal model, so the projectile is still moving. Its acceleration remains −g vertically; gravity does not switch off at the apex. Setting the total velocity or acceleration to zero here confuses a component condition with a vector condition.

The time to the apex follows from 0 = u_y − gt, giving t_apex = u_y/g. The vertical rise follows from v_y² = u_y² − 2gΔy, giving Δy_max = u_y²/(2g). For u_y = 10.0 m s⁻¹, these are 1.02 s and 5.10 m. A platform 12.0 m above ground therefore gives a highest height of 17.1 m above ground, even though the rise from launch is only 5.10 m.

At equal heights on opposite sides of the apex, ideal vertical velocities have equal magnitudes and opposite signs. The horizontal components are identical. Consequently the speeds are equal, but the velocity vectors are not. The times from the apex to those equal-height positions are also equal; this symmetry does not imply equal ascent and descent durations when the final landing is lower than launch.

A useful graphical check is to plot v_y against t. It is a straight line of slope −g crossing zero at the apex. The area under the line is vertical displacement; positive area during ascent is offset by negative area on descent. A plot of speed against time is different because speed combines both perpendicular components and remains positive for a launch with nonzero horizontal velocity.

4. Calculate and communicate the impact velocity

A request for velocity requires both magnitude and direction. At landing, calculate v_x and v_y separately, then reconstruct the vector. For the platform example, v_x = 17.3205 m s⁻¹ and v_y = 10.0 − 9.80 × 2.8886 = −18.308 m s⁻¹. Thus the impact speed is √(17.3205² + 18.308²) = 25.2 m s⁻¹.

The direction below horizontal is tan⁻¹(18.308/17.3205) = 46.6°. State the result as 25.2 m s⁻¹ at 46.6° below the forward horizontal, or give the signed component vector if requested. Writing merely −25.2 m s⁻¹ does not define a two-dimensional direction. When using an inverse tangent, identify the quadrant from the component signs rather than relying on the calculator's principal-angle output.

Energy provides an independent check of the speed. With drag neglected, ½mv² = ½mu² + mg(12.0), so v² = 20.0² + 2 × 9.80 × 12.0 = 635.2 m² s⁻², giving the same 25.2 m s⁻¹. Energy alone does not provide the impact direction, which is why the component calculation is still necessary.

Keep extra digits in intermediate values and round the final quantity to the precision requested. Rounding both components too early can noticeably affect a reconstructed angle or a small difference. Include units in the substituted or final expression, and distinguish a signed vertical velocity from a positive speed. These are separate pieces of physical information that can earn separate marks in a multipart response.

5. Use a trajectory equation carefully

When a question specifies a horizontal position rather than a time, eliminate t using t = x/u_x. For u_x ≠ 0, the ideal trajectory is y = x tan θ − gx²/(2u² cos²θ). This is a parabola in the chosen coordinate system. It gives vertical displacement at horizontal position x; add the launch height if the required answer is height above a ground datum.

Suppose a ball is launched at 18.0 m s⁻¹ and 40.0° from 1.50 m above ground. At x = 12.0 m, u_x = 13.79 m s⁻¹ and t = 0.870 s. Its rise from launch is (18.0 sin 40.0°)(0.870) − 4.90(0.870)² = 6.36 m, so its height is 7.86 m. A 7.00 m obstacle would be cleared by 0.86 m in this ideal model. Reporting 6.36 m as the ground height would omit the datum shift.

For equal launch and landing heights, substituting y = 0 yields the range R = u² sin 2θ/g. At a fixed speed, complementary launch angles have equal ideal ranges and 45° maximises range. Neither conclusion can be transferred automatically to a launch from a cliff, a moving target or significant drag. The range formula has assumptions, not universal authority.

A trajectory equation cannot describe a vertical launch because division by u_x is then undefined. Use the time equations instead. Similarly, if a question asks for the instant of impact on a sloping surface, equate the trajectory to the surface equation and select the physically relevant intersection. Record why a root is valid rather than treating every algebraic root as an attainable event.

6. Turn a solution into an assessable argument

Scope note. The uncertainty calculation below is an optional first-order extension, included as a supplied method rather than a required memorised subject-outline formula.

A strong projectile response exposes the reasoning that connects the diagram to the answer. State the axes and initial components; choose the equation that matches the event; substitute with correct signs; solve in the physical time domain; and then calculate the requested component, magnitude or direction. If the question says “show”, reproduce the stated value from valid working rather than substituting that value back as though it were independent evidence.

In a launch-speed experiment, a horizontal launch from measured height h provides t = √(2h/g), and measured range R then gives u = R√(g/(2h)). This relation assumes the projectile leaves horizontally, the landing level is known and drag is negligible. Measuring from the launcher housing rather than the release point introduces a systematic range offset. Repeating launches reduces uncertainty in the mean range but does not repair that offset.

For small independent measurement uncertainties, a conservative first-order estimate gives Δu/u ≈ ΔR/R + ½Δh/h. If R = 2.00 ± 0.02 m and h = 0.800 ± 0.005 m, the estimate is 0.010 + 0.00313 ≈ 1.31%. This is an uncertainty estimate based on the stated approach, not a proof that the ideal model is correct. A consistent shortfall in observed range may reflect drag or a downward launch angle rather than random scatter.

When comparing model and measurement, discuss a specific assumption and its predicted effect. Drag opposing forward motion reduces horizontal speed, so x = u_xt becomes an over-simplification. A slight downward launch angle reduces flight time compared with a horizontal release at the same height. Distinguishing these mechanisms demonstrates understanding beyond saying that an experiment contains “human error”.

7. Worked investigation: a horizontal launch and an obstacle

Problem. A steel ball leaves a level bench 45.0 m above level ground with horizontal speed 18.0 m s⁻¹. A vertical wall begins 36.0 m horizontally from the launch point and rises 26.0 m above the same ground. Treat the ball as a particle, ignore air resistance and use g = 9.80 m s⁻². Determine whether the ball clears the wall, where it first reaches ground, and its velocity just before impact. A correct solution has to use one common clock for horizontal and vertical motion; it must not invent a horizontal acceleration merely because gravity acts vertically.

Choose the launch point as x = 0 and ground as y = 0, with upward positive. Initially x₀ = 0, y₀ = 45.0 m, vₓ = +18.0 m s⁻¹ and vᵧ₀ = 0. The component equations are x = 18.0t and y = 45.0 − 4.90t². The ball reaches the wall plane when t = 36.0/18.0 = 2.00 s. At this instant y = 45.0 − 4.90(2.00)² = 25.4 m. The top of the wall is 26.0 m high, so the modelled ball is 0.60 m below the top. It strikes the wall in this particle model; a ground-range answer is therefore a counterfactual “if the wall were removed”, rather than its actual path after collision. This qualification matters: the mathematical trajectory can be extended past an obstacle, but the physical trajectory cannot.

With the wall removed, set the ground landing condition y = 0, giving 0 = 45.0 − 4.90t². The physically relevant positive root is t = √(90.0/9.80) = 3.03046 s. The negative mathematical root describes a time before launch and is rejected. Horizontal range is x = 18.0(3.03046) = 54.5482 m, or 54.5 m to three significant figures. At that instant the velocity components are vₓ = 18.0 m s⁻¹ and vᵧ = −9.80(3.03046) = −29.6985 m s⁻¹. Combining perpendicular components gives speed √(18.0² + 29.6985²) = 34.7275 m s⁻¹ and direction arctan(29.6985/18.0) = 58.8° below the horizontal. The minus sign on vᵧ locates the quadrant; speed itself cannot be negative.

Check and interpretation. At the wall time the vertical displacement is −19.6 m, not −45.0 m; the ball has not yet reached ground. At landing the horizontal acceleration is zero, so vₓ remains the launch value. The two times are different because a wall crossing is determined by a given x whereas ground impact is determined by a given y. If the wall height were instead 25.0 m, the same projectile would clear it by 0.40 m, subject to ball radius and measurement uncertainty. An exam response should state both the arithmetic and the physical decision, with metres attached to clearances and m s⁻¹ attached to velocity components.

Follow-up design calculation. Suppose the wall cannot be moved but the horizontal launch speed can be adjusted, still with zero initial vertical speed. What minimum speed permits the ball’s centre to reach the wall plane at or above 26.0 m? At the limiting case y = 26.0, solve 26.0 = 45.0 − 4.90t², giving t = √(19.0/4.90) = 1.96915 s. Since the wall is 36.0 m away, vₓ,min = 36.0/1.96915 = 18.2820 m s⁻¹. A setting of 18.3 m s⁻¹ would just satisfy the ideal particle model, whereas the original 18.0 m s⁻¹ did not. In a real setup a finite-radius ball requires its entire surface above the obstacle, and a margin is prudent because 26.0 m and 45.0 m are measured, rounded heights. The inverse problem demonstrates that gravity controls the allowed arrival time while launch speed controls the horizontal distance achieved in that time.

A student might instead insert x = 36 directly into y = 45 − 4.9x² and obtain a nonsensical large negative height. That expression uses seconds for its squared variable; x is measured in metres. To eliminate time properly, substitute t = x/vₓ, obtaining y(x) = 45 − 4.90(x/vₓ)². This trajectory form makes dimensional consistency visible: x/vₓ has units s. It also shows that raising vₓ makes the same horizontal position arrive sooner and therefore reduces the vertical drop. Do not claim horizontal velocity changes the time to fall a specified vertical distance when the vertical starting state is fixed; the changed outcome is whether the wall is reached before or after that vertical drop.

8. Worked inverse design: launching upward from a raised platform

Problem. A demonstration launcher stands 12.0 m above a field and fires a ball at 25.0 m s⁻¹, 35.0° above horizontal. Predict its maximum height above the field, flight duration, horizontal landing point and impact speed, using g = 9.80 m s⁻² and negligible drag. Then decide whether a target whose centre is 70.0 m from the launcher and at field level lies short of or beyond the predicted landing point. The launch angle is measured above the horizontal, so both initial components are positive.

Resolve first: vₓ₀ = 25.0 cos 35.0° = 20.4788 m s⁻¹ and vᵧ₀ = 25.0 sin 35.0° = 14.3394 m s⁻¹. Use field level for y = 0 and upward positive, so y(t) = 12.0 + 14.3394t − 4.90t²; horizontal position is x(t) = 20.4788t. The apex occurs when vᵧ = 14.3394 − 9.80t = 0, at t = 1.46321 s. The vertical gain above the launcher is vᵧ₀²/(2g) = 10.4908 m, giving maximum field-relative height 12.0 + 10.4908 = 22.4908 m. Do not call 10.4908 m the height above the field: it is only the rise from an already elevated launcher.

For landing, solve 0 = 12.0 + 14.3394t − 4.90t². The positive quadratic root is t = [14.3394 + √(14.3394² + 2(9.80)(12.0))]/9.80 = 3.60562 s. The other root is negative and not a post-launch event. Horizontal range is 20.4788(3.60562) = 73.8388 m. The field-level target centre at 70.0 m lies 3.84 m short of this ideal landing point. If the target has a finite radius, whether it is hit requires its dimensions, not just the centre distance. The impact components are vₓ = +20.4788 and vᵧ = 14.3394 − 9.80(3.60562) = −20.9957 m s⁻¹. Therefore the impact speed is √(20.4788² + 20.9957²) = 29.3292 m s⁻¹, greater than 25.0 m s⁻¹ because gravitational potential energy is lower at the field than at the launch platform.

Independent energy check. Without drag, v² = u² + 2gΔh for a fall Δh = 12.0 m. Thus v² = 25.0² + 2(9.80)(12.0) = 860.2 m² s⁻² and v = √860.2 = 29.3292 m s⁻¹, confirming the component result without solving the trajectory again. A frequent wrong shortcut is to double the apex time for the full flight. That works only when launch and landing heights are equal; here 2(1.46321) = 2.92642 s would miss the extra descent from launch height to field. State the height datum, root choice and model assumptions when reporting any projectile range.

Inverse target check. Suppose a launcher technician wants this same 35.0° angle and 12.0 m launch height, but the ideal trajectory must reach field level exactly 70.0 m away. What initial speed would the model require? Eliminate time using t = x/(u cos 35°) in y = 12.0 + (u sin 35°)t − 4.90t². At x = 70.0 m this becomes 0 = 12.0 + 70.0 tan 35° − 9.80(70.0)²/[2u² cos²35°]. Rearranging gives u = √{9.80(70.0)²/[2cos²35°(12.0 + 70.0tan35°)]} = 24.2167 m s⁻¹. This is lower than 25.0 m s⁻¹, consistent with the original shot overshooting to 73.84 m. The denominator is positive here, so the square root is physically meaningful. This inverse result is a predicted setting, not evidence that the real launcher will hit the target when air resistance and release-height uncertainty are present.

If the technician measures an actual landing point short of 70.0 m at that setting, it does not follow that the component equations were algebraically wrong. Drag acts backward through the trajectory and can shorten range; the launch speed or angle might also differ from the setting. Repeat shots, measure the actual initial velocity and release height, and compare model residuals. A clear final response separates the ideal prediction, a measured outcome and a causal explanation. In calculations, keep full precision in sin 35° and cos 35° until the final displayed speed; prematurely using 0.57 and 0.82 can shift a borderline target decision.

The impact-speed energy check is especially useful if the landing time has been rounded: it depends only on the 12.0 m height difference and initial speed, so it should agree with the component calculation to the precision of the inputs. If the two answers differ materially, recheck the sign of vertical displacement before blaming rounding.