Analyse the motion of projectiles by resolving the motion into horizontal and vertical components, making the following assumptions: a constant vertical acceleration due to gravity, zero air resistance

Resolve the initial velocity into components.

$v_{0x} = 25 \cos 40° = 19.15$ m/s.
$v_{0y} = 25 \sin 40° = 16.07$ m/s.

Maximum height occurs when $v_y = 0$. Using $v_y^2 = v_{0y}^2 - 2gh$ with $v_y = 0$:

$h = \frac{v_{0y}^2}{2g} = \frac{16.07^2}{2 \times 9.8} = 13.2$ m.

Range. Total time of flight to return to ground level: $t = \frac{2 v_{0y}}{g} = \frac{2 \times 16.07}{9.8} = 3.28$ s.

Range $R = v_{0x} t = 19.15 \times 3.28 = 62.8$ m.

Markers reward clear resolution of components, correct use of SUVAT on each axis, and answers stated with units and to two or three significant figures.

Horizontal and vertical components are independent. Initial vertical velocity is zero (thrown horizontally).

Time of flight. Using $y = \frac{1}{2} g t^2$ with $y = 45$ m:

$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 45}{9.8}} = 3.03$ s.

Horizontal distance. Horizontal velocity is constant.

$x = v_x t = 12 \times 3.03 = 36.4$ m.

Markers expect explicit statement that $v_{0y} = 0$, correct identification that the only acceleration is gravity, and final answers with units.

A projectile is an object in flight that is subject only to the force of gravity (its weight); no other force acts on it once launched.

The two simplifying assumptions are: (1) the vertical acceleration is constant, $a_y = -g = -9.8\ \text{m/s}^2$, and (2) air resistance is ignored, so horizontal velocity never changes.

Marks: one for the definition (subject only to gravity), one for both assumptions named (constant $g$ and zero air resistance).

$v_{0x} = v_0 \cos\theta = 8.0 \cos 55° = 8.0 \times 0.5736 = 4.6$ m/s.

$v_{0y} = v_0 \sin\theta = 8.0 \sin 55° = 8.0 \times 0.8192 = 6.6$ m/s.

Marks: one for the correct formula for each component, one for correct substitution, one for both answers to two significant figures with the unit m/s.

Thrown horizontally, so $v_{0y} = 0$. Using $y = \tfrac{1}{2} g t^2$ with $y = 20.0$ m:

$t = \sqrt{\dfrac{2y}{g}} = \sqrt{\dfrac{2 \times 20.0}{9.8}} = \sqrt{4.08} = 2.0$ s.

Horizontal velocity is constant: $x = v_x t = 15 \times 2.0 = 30$ m.

Marks: one for stating $v_{0y} = 0$ and the correct vertical SUVAT equation, one for $t = 2.0$ s, one for $x = 30$ m with the unit.

(a) Maximum height. $v_{0y} = 45 \sin 30° = 22.5$ m/s. At the peak $v_y = 0$, so

$h = \dfrac{v_{0y}^2}{2g} = \dfrac{22.5^2}{2 \times 9.8} = \dfrac{506.25}{19.6} = 25.8$ m.

(b) Range. $R = \dfrac{v_0^2 \sin(2\theta)}{g} = \dfrac{45^2 \sin 60°}{9.8} = \dfrac{2025 \times 0.8660}{9.8} = 179$ m.

Marks: one for $h = v_{0y}^2/(2g)$ with substitution, one for $h = 25.8$ m, one for the range formula with substitution, one for $R = 179$ m with the unit.

Rearrange the range formula for $v_0$: $R = \dfrac{v_0^2 \sin(2\theta)}{g} \Rightarrow v_0 = \sqrt{\dfrac{Rg}{\sin(2\theta)}}$.

$v_0 = \sqrt{\dfrac{65 \times 9.8}{\sin 84°}} = \sqrt{\dfrac{637}{0.9945}} = \sqrt{640.5} = 25.3$ m/s.

Marks: one for correctly rearranging $R = v_0^2 \sin(2\theta)/g$ for $v_0$, one for substituting $2\theta = 84°$ correctly (not $42°$), one for $v_0 = 25.3$ m/s with the unit.

(a) Both $\theta = 30°$ and $\theta = 60°$ give $R \approx 35.3$ m on the graph. This illustrates that complementary angles (angles that sum to $90°$) produce the same range for a given launch speed.

(b) The graph peaks at $\theta = 45°$. At this angle, $R_{\max} = \dfrac{v_0^2 \sin(2\theta)}{g} = \dfrac{20^2 \sin 90°}{9.8} = \dfrac{400 \times 1}{9.8} = 40.8$ m.

(c) $R = \dfrac{v_0^2 \sin(2\theta)}{g}$ depends on $\theta$ only through $\sin(2\theta)$. Since $\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90° - \theta))$, an angle $\theta$ and its complement $90° - \theta$ always give the same $\sin(2\theta)$ value and hence the same range, which makes the curve symmetric about the peak at $\theta = 45°$ (where $\sin(2\theta) = 1$, its maximum possible value).

Marks: one for reading $R \approx 35.3$ m at both angles from the graph, one for stating the complementary-angle relationship, one for identifying $\theta = 45°$, one for $R_{\max} = 40.8$ m, one for the $\sin(2\theta) = \sin(2(90° - \theta))$ symmetry argument.

Band-6 plan. (1) State the standard result: $45°$ maximises range when launch and landing heights are equal, and derive it from $R = v_0^2 \sin(2\theta)/g$. (2) Explain why $\sin(2\theta)$ is maximised at $\theta = 45°$. (3) Analyse the shoulder-height case: with a launch height above the landing height, the ball spends extra time falling, so the optimal angle is actually LESS than $45°$ (a flatter throw uses more of $v_0$ for horizontal distance during that extra fall time). (4) Conclude: the claim is correct only for equal launch/landing heights, not in general.

Model answer. For a projectile launched and landing at the same height, the range is $R = \dfrac{v_0^2 \sin(2\theta)}{g}$. Since $v_0$ and $g$ are fixed, $R$ is maximised when $\sin(2\theta)$ is largest. The sine function has its maximum value of $1$ at an angle of $90°$, so $2\theta = 90°$, giving $\theta = 45°$. In this equal-height case the fielder's claim is correct: no other angle gives a longer range for the same launch speed.

However, a fielder throws from shoulder height (typically $\sim 1.8$ m) to a catch made near ground level, so the landing point is below the launch point. The ball is now in the air for longer than the level-ground formula predicts, because it must fall the extra $1.8$ m below the launch height as well as return to launch height. A flatter angle (less than $45°$) gives a larger horizontal component $v_0 \cos\theta$ while still keeping the ball airborne for this longer, extended flight time, so the horizontal distance $x = v_{0x} t$ is increased even though $\sin(2\theta)$ alone would favour $45°$. Full projectile analysis (solving $-1.8 = v_0 \sin\theta \, t - \tfrac{1}{2} g t^2$ for $t$, then maximising $x = v_0 \cos\theta \, t$ over $\theta$) shows the optimal angle drops below $45°$, more so as the launch height or the speed decreases relative to the drop.

The claim is therefore only correct when the launch and landing heights are equal. When the ball is released above the level it lands at, as in a fielder's throw, the range-maximising angle is somewhat less than $45°$, because the ball spends extra unaccounted time falling that a flatter, faster horizontal component exploits.

Marker's note: the top band derives $\theta = 45°$ from the range formula (not just quotes it), correctly identifies that unequal launch/landing height breaks the symmetry, and reasons (without necessarily completing the full calculus) that the optimal angle shifts below $45°$ when launching from above the landing point. A response that states "$45°$ is always best" without addressing the height difference caps in the middle band.

Band-6 plan. (1) State the zero-air-resistance assumption and why it is used (it makes the horizontal and vertical motions independently solvable with simple SUVAT). (2) Name a real sporting projectile and note that real air resistance (drag, proportional to speed squared, and for spinning balls the Magnus effect) is not actually zero. (3) Explain the qualitative effect on the trajectory (asymmetric path, reduced range, earlier and steeper descent). (4) Explain the effect on the optimal angle (drag typically shifts the optimal launch angle below $45°$). (5) Conclude with a judgement of validity: a reasonable simplifying model for HSC calculation purposes and slow/dense projectiles, but increasingly inaccurate for fast, light projectiles.

Model answer. The zero-air-resistance assumption lets horizontal and vertical motion be treated completely independently: horizontal velocity is constant and vertical motion obeys simple SUVAT equations with constant acceleration $g$. This produces the clean parabolic trajectory and the closed-form results $R = v_0^2\sin(2\theta)/g$ and $h = v_0^2\sin^2\theta/(2g)$ used throughout Module 5.

In reality, no projectile moves through a true vacuum. A golf ball, for example, experiences aerodynamic drag proportional to the square of its speed, which is largest at launch (when speed is highest) and removes kinetic energy throughout the flight. This makes the real trajectory asymmetric: the ball decelerates more on the way up than the idealised parabola predicts, so the ascending part of the path is flatter than the descending part, and the ball falls more steeply near the end of its flight than it rose at the start.

The practical consequences are a shorter range than the drag-free formula predicts, sometimes drastically so (a driven golf ball's true range is far below the $v_0^2\sin(2\theta)/g$ prediction for its initial speed), and a shift in the optimal launch angle. Because drag removes more energy from a projectile that spends longer in the air, the range-maximising angle for a draggy projectile is typically well below $45°$, often closer to $30$-$40°$ for a fast golf shot, since a flatter trajectory reduces the time (and hence the drag losses) at the cost of the extra height a $45°$ shot would gain. A golf ball's backspin also generates lift via the Magnus effect, partially offsetting drag and letting real driven balls carry farther, and at a lower optimal angle, than a spinless drag calculation alone would predict.

Overall, the zero-air-resistance assumption is a reasonable and necessary simplification for introductory HSC calculations. It gives the correct qualitative shape of the motion and is a fair approximation for slow, dense, compact projectiles (a shot-put, a dropped stone) where drag forces are small compared to weight. It becomes markedly less valid for fast, light, or spin-affected projectiles such as a golf ball, cricket ball or javelin, where real ranges and optimal angles depart significantly from the idealised formulas, so any quoted HSC answer should be understood as a model, not a literal prediction of real flight.

Marker's note: the top band explains WHY the assumption simplifies the mathematics (independent axes, closed-form SUVAT), names a specific real projectile with a plausible qualitative drag effect, links drag to BOTH a reduced range and a lowered optimal angle (not just "it goes less far"), and closes with an explicit judgement of validity across different projectile types. A response that only asserts "air resistance makes things less accurate" without these mechanisms caps in the middle band.