Physics • Year 12 • Module 5 • Lesson 3
Problem Solving: Projectiles
Apply the GIVEN→FIND→METHOD→ANSWER protocol to real-data graph reading, cause-and-effect reasoning, step-sequencing, and prediction tasks involving projectile motion.
1. Interpret the graph, vertical velocity of a projectile
The graph below shows the vertical component of velocity ($v_y$) versus time for a projectile launched from ground level at $22\ \text{m/s}$ and $42°$ above horizontal. 7 marks
Figure 1. Vertical velocity vs time for a projectile: $v = 22\ \text{m/s}$, $\theta = 42°$, launched from ground level. $g = 9.8\ \text{m/s}^2$. Illustrative data.
1.1 Describe the shape of the graph and explain what it tells you about the acceleration acting on the projectile. 2 marks
1.2 From the graph, estimate the time at which the projectile reaches maximum height. Explain how you identified this point. 2 marks
1.3 Calculate the maximum height reached by the projectile. Use $v_y^2 = u_y^2 - 2gh_{\text{max}}$ and show all working. 3 marks
2. Sequence the steps, solving a projectile-over-a-wall problem
A projectile is launched from a cliff at 18 m/s, 35° above horizontal. You need to determine whether it clears a wall 25 m away. The eight steps below are in the wrong order. Write the correct sequence (1–8) in the “Order” column. 8 marks (1 per correct position)
| Order | Step |
|---|---|
| Calculate vertical displacement using $s_y = v_y t + \frac{1}{2}(-g)t^2$. | |
| State the FIND: the vertical position of the ball when it reaches the wall. | |
| Compare the calculated height with the wall height to answer the question. | |
| Resolve the launch velocity into components: $v_x = 18\cos 35°$, $v_y = 18\sin 35°$. | |
| Draw a labelled diagram showing cliff, wall, horizontal distance, and vertical heights. | |
| Add the cliff height to the vertical displacement to find height above sea level. | |
| List GIVEN quantities: $v = 18\ \text{m/s}$, $\theta = 35°$, cliff height, wall distance, wall height. | |
| Find the time to reach the wall using $t = s_x / v_x$. |
3. Cause-and-effect chain, launching at a steeper angle
A javelin thrower increases their launch angle from 35° to 55° while keeping the launch speed constant. Complete the cause-and-effect chain below. Write the effect of each change in the empty boxes. 5 marks (1 per effect)
4. Predict and justify, Olympic long jump
An Olympic long jumper leaves the take-off board at $9.6\ \text{m/s}$ at an angle of $20°$ above horizontal. The landing pit surface is at the same height as the take-off board. 5 marks
4.1 Predict the horizontal distance (range) of the jump. Use the range equation and show your working using the GIVEN→FIND→METHOD→ANSWER protocol. Take $g = 9.8\ \text{m/s}^2$. 3 marks
4.2 The world record long jump is 8.95 m (Mike Powell, 1991). Justify whether the range equation predicts a realistic result, and identify one key assumption that limits its accuracy in modelling a real long jump. 2 marks
Q1.1, Shape and acceleration (2 marks)
The graph is a straight line with a negative slope [1]. This indicates a constant acceleration in the downward direction: the vertical velocity decreases at a uniform rate of $9.8\ \text{m/s}$ per second due to gravity. The acceleration ($g = 9.8\ \text{m/s}^2$ downward) is constant throughout the flight [1].
Q1.2, Time at maximum height (2 marks)
Maximum height occurs at approximately $t = 1.5\ \text{s}$ [1]. At this point the vertical velocity is zero, the line crosses the time axis ($v_y = 0$). When vertical velocity is zero, the projectile has momentarily stopped rising and is about to begin descending, which is the definition of maximum height [1].
Q1.3, Maximum height calculation (3 marks)
$u_y = 22\sin 42° = 22 \times 0.669 = 14.72\ \text{m/s}$ [1]
$v_y^2 = u_y^2 - 2gh_{\text{max}} \Rightarrow 0 = 14.72^2 - 2(9.8)h_{\text{max}}$ [1]
$h_{\text{max}} = 14.72^2 / (2 \times 9.8) = 216.7 / 19.6 = 11.1\ \text{m}$ [1]
Q2, Sequence the steps (correct order)
1: Draw a labelled diagram. 2: List GIVEN quantities. 3: State the FIND. 4: Resolve velocity components. 5: Find time to reach the wall ($t = s_x/v_x$). 6: Calculate vertical displacement ($s_y$). 7: Add cliff height to get height above sea level. 8: Compare with wall height.
Q3, Cause-and-effect chain (5 marks)
Effect 1: Decreases (cos 55° < cos 35°, so $v_x$ is smaller at 55°) [1].
Effect 2: Increases (sin 55° > sin 35°, so $v_y$ is larger at 55°) [1].
Effect 3: Increases (larger $v_y$ means the projectile takes longer to decelerate to zero) [1].
Effect 4: Increases (higher vertical velocity means greater maximum height: $h_{\text{max}} = v_y^2/2g$) [1].
Overall outcome: The range remains the same as the 35° launch (because $\sin 2\times 35° = \sin 70° = \sin 110° = \sin 2\times 55°$, so the range equation gives the identical value). The flight path is higher but shorter horizontally, and the two effects cancel to produce the same range [1].
Q4.1, Long jump range (3 marks)
GIVEN: $v = 9.6\ \text{m/s}$, $\theta = 20°$, level ground, $g = 9.8\ \text{m/s}^2$ [1]
FIND: Range $R$
METHOD + ANSWER: $R = v^2\sin(2\theta)/g = 9.6^2 \times \sin(40°)/9.8 = 92.16 \times 0.643/9.8 = 59.3/9.8 = 6.05\ \text{m}$ [1]
The predicted range is 6.05 m (3 s.f.) [1 for correct final answer with units].
Q4.2, Evaluation (2 marks)
The prediction of 6.05 m is realistic, it is within the range of competitive long jump distances (typical 7–9 m for elite athletes, lower for a modest launch speed of 9.6 m/s) [1]. A key assumption limiting accuracy is that the landing position is at exactly the same height as the take-off board; in reality, athletes land in a sand pit and the effective landing point (heel contact) is slightly below the board height, increasing the measured distance. Air resistance is also neglected [1].