A projectile is launched from level ground with speed 30ms−1. It must pass through the point (40,15), where coordinates are in metres. Take g=10ms−2 and ignore air resistance. Determine the two possible acute launch angles, to the nearest 0.1∘.
For launch angle θ, x=30cosθt,y=30sinθt−5t2. Eliminate t=x/(30cosθ): y=xtanθ−180x2sec2θ. At (40,15), let q=tanθ and use sec2θ=1+q2: 15=40q−980(1+q2). Rearranging gives 16q2−72q+43=0, so q=49±38. Taking the inverse tangent of both positive roots gives θ≈35.3∘ or 75.2∘.
Forms the projectile model.
[2 marks]
Eliminates time.
[1 mark]
Forms the quadratic in tanθ.
[1 mark]
Determines both tangent values.
[1 mark]
Determines both launch angles.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.