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Vector calculus: projectile interception — Question 167

Original QCE Vault practice · 7 marks

Q167 · Practice questionTechnology-freeComplex unfamiliar7 marks

QUESTION 167 (7 marks)

A projectile is launched from the origin at t=0t=0 with velocity 8i+12j8\mathbf i+12\mathbf j metres per second and constant acceleration −4j-4\mathbf j metres per second squared. A target moves with position vector
rT(t)=(30+2t)i+10j,t≥0.\mathbf r_T(t)=(30+2t)\mathbf i+10\mathbf j,\qquad t\ge0.
The projectile model applies until it first returns to ground level y=0y=0. The diagram shows its path and the target's horizontal path.
Projectile path y=1.5x-x squared/32 from (0,0) to (48,0), and target path y=10 for x at least 30. Initial target position T0=(30,10) and its rightward direction are marked.
a)
Determine the projectile's position vector and its flight-time interval.
[2 marks]
b)
Determine whether the projectile hits the moving target. If it does, give the time and position.
[3 marks]
c)
Determine the maximum height and state whether the projectile is ascending or descending at interception.
[2 marks]
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