Projectile Range from a Bench Edge: When 45° Is Not the Maximum
In a Grade 11 mechanics practical, groups use a spring-loaded launcher fixed to a bench edge 0.90 m above the floor. The launcher gives a ball an initial speed of 4.2 m/s at selected angles to the horizontal. Students measure horizontal range to the first floor bounce. The class prediction, from textbook projectile theory, is that range is greatest at 45°. Group A obtains: 30°, 2.48 m; 45°, 2.42 m; 60°, 1.96 m. A student checks the 45° result using R = v² sin 2θ / g = (4.2)² / 9.81 = 1.80 m and claims the apparatus is faulty because the measured range is much larger. Another student says the 30° result proves the teacher's formula is wrong. The teacher has one double lesson and a second launcher available. The class is preparing for an examination question on projectile motion from a height.
1. Explain, with a calculation, why the 45° launch need not give the maximum range in this arrangement. State whether Group A's data are consistent with a constant-acceleration projectile model.
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The ball leaves at height h = 0.90 m and lands at y = 0, so the vertical displacement is -0.90 m. Using s = ut + ½at² gives -0.90 = v sinθ t - ½gt², so t = [v sinθ + √(v² sin²θ + 2gh)] / g, and R = v cosθ t. With v = 4.2 m/s and g = 9.81 m/s²:
- 30°: t = 0.693 s, R = 2.52 m
- 45°: t = 0.827 s, R = 2.46 m
- 60°: t = 0.937 s, R = 1.97 m
The measured values 2.48, 2.42 and 1.96 m are within about 2% of these predictions, so the data are consistent with the constant-acceleration model. The 45° angle maximises range only when launch and landing heights are equal. Here the extra 0.90 m fall gives lower-angle launches more time to travel horizontally, so the maximum is nearer 30–35°.
2. A student used R = v² sin 2θ / g. Identify the assumption in that equation and explain how the error affects the student's conclusion about the apparatus.
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The equation R = v² sin 2θ / g is derived for a projectile that lands at the same height from which it is launched. It ignores the extra 0.90 m fall in this experiment. Using it at 45° gives 1.80 m, below the measured 2.42 m; the discrepancy is not evidence of a faulty launcher but of an invalid assumption. The student's conclusion is therefore unsupported. The correct test is to compare the measured range with the height-corrected model, or to raise the landing surface to the launch height and repeat the measurements. This distinction matters because a model can be reliable within its conditions and still give poor predictions outside them.
3. The teacher wants to use the discrepancy as a formative assessment opportunity. Describe two specific instructional moves, including how to handle the student who claims the formula is wrong.
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First, have students derive the height-corrected time equation from the suvat equations rather than substitute a remembered formula. This makes the level-landing assumption visible and gives students ownership of the correction. Second, use a predict–observe–explain cycle: ask groups to predict the angle of maximum range for h = 0 and for h = 0.90 m, then plot class data with the model curve. For the student who says the formula is wrong, respond that the formula is a model with conditions; ask them to identify which condition fails. This validates their observation while correcting the overgeneralisation. Record the episode as evidence against the misconception that equations are unconditionally true, and use a similar item in the next assessment.