Projectile Range with a Fence Constraint
A ball is launched from ground level at 18 m/s at an angle of 35° above the horizontal. A vertical fence of height 4.2 m stands 14 m from the launch point, measured horizontally. Take g = 9.8 m/s² and ignore air resistance. The school's marking guide awards method marks for resolving velocity, forming the trajectory equation, and interpreting the discriminant, and a final accuracy mark for the conclusion. A student writes the horizontal position as x = 18t and the vertical position as y = 18t − 4.9t², then substitutes t = x/18 without resolving components. A second student correctly resolves the components but claims the ball clears the fence because the maximum height exceeds 4.2 m.
1. Identify the error in each student's reasoning and determine, with full working, whether the ball clears the fence.
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Student 1's error: the initial speed was not resolved into components. Correctly, uₓ = 18cos35° ≈ 14.74 m/s and u_y = 18sin35° ≈ 10.32 m/s. Using the full speed for both is a dimensionally plausible but physically wrong substitution.
Student 2's error: maximum height being above 4.2 m does not guarantee clearance at x = 14 m; the ball may already be descending below the fence height at that horizontal position. Clearance must be tested at the fence's x-coordinate, not at the apex.
Correct test: time to reach x = 14 m is t = 14/14.74 ≈ 0.9497 s. Then y = 10.32(0.9497) − 4.9(0.9497)² ≈ 9.80 − 4.42 ≈ 5.38 m. Since 5.38 m > 4.2 m, the ball clears the fence by about 1.2 m. (Equivalently, substitute x into y = x tan35° − gx²/(2u²cos²35°) and evaluate at x = 14.)
2. The launch angle can be varied while keeping the speed at 18 m/s. Determine the range of angles for which the ball clears the fence, and explain why the boundary angles are found from a quadratic in tan θ.
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Write the trajectory as y = x tanθ − (gx²)/(2u²cos²θ). Using 1/cos²θ = 1 + tan²θ and substituting x = 14, u = 18, g = 9.8:
y = 14 tanθ − (9.8·196)/(2·324)(1 + tan²θ) = 14 tanθ − 2.9642(1 + tan²θ).
Set y = 4.2 (grazing the fence): 2.9642 tan²θ − 14 tanθ + 7.1642 = 0. Solving, tanθ = [14 ± √(196 − 84.94)]/5.9284 = [14 ± 10.540]/5.9284, giving tanθ ≈ 4.132 or 0.5835, so θ ≈ 76.4° or 30.3°.
The quadratic in tanθ arises because the trajectory equation is quadratic in tanθ once the identity 1/cos²θ = 1 + tan²θ is applied. Between the two roots the parabola lies above the fence line; outside it falls short. Hence the ball clears the fence for 30.3° < θ < 76.4° (approximately). The upper bound reflects that a very steep launch descends steeply and can strike the fence on the way down.