When the Discriminant Lies About a Real-World Model
A projectile is launched from ground level with speed 20 m/s at angle θ to the horizontal. A wall of height 8 m stands 15 m away. A student sets up the trajectory y = x tanθ − gx²/(2v²cos²θ) with g = 9.8, substitutes x = 15, y = 8, and forms a quadratic in tanθ. They compute the discriminant and find it negative, concluding the projectile can never clear the wall. However, the teacher notes the student used v = 20 but the launcher was actually rated at 20 m/s only at ground level; air resistance is ignored, and the wall is thin. The student's algebra is correct but the conclusion is suspect because they solved for tanθ rather than checking the maximum height.
1. Show the correct quadratic in tanθ for clearing the wall at exactly 8 m, and determine whether a real launch angle exists.
Read the worked answer
Using y = x tanθ − (g x²)/(2v²)(1 + tan²θ) with x = 15, y = 8, v = 20, g = 9.8:
8 = 15T − (9.8·225)/(2·400)(1 + T²), where T = tanθ.
(9.8·225)/800 = 2205/800 = 2.75625.
So 8 = 15T − 2.75625 − 2.75625T².
Rearrange: 2.75625T² − 15T + 10.75625 = 0.
Discriminant = 225 − 4(2.75625)(10.75625) = 225 − 118.58 ≈ 106.42 > 0.
So two real angles exist; the student's negative discriminant came from a sign or arithmetic slip, not from physics. The judgement call is to distrust a surprising conclusion and re-derive.
2. Find the launch angles that make the projectile pass exactly through the top of the wall, and explain which is preferable.
Read the worked answer
T = [15 ± √106.42]/(2·2.75625) = [15 ± 10.316]/5.5125.
T₁ = 25.316/5.5125 ≈ 4.593, T₂ = 4.684/5.5125 ≈ 0.850.
θ₁ = arctan(4.593) ≈ 77.7°, θ₂ = arctan(0.850) ≈ 40.4°.
The low angle (≈40.4°) is preferable: it reaches the wall sooner and with a shallower descent, and the higher launch is steep and sensitive to small errors in speed. Both are valid mathematically; the modelling judgement is to choose the robust one.