2022 HSC Mathematics Extension 1
Question 14(c):
Vector motion
A projectile is launched from the origin with speed at angle to the horizontal to hit a target that starts at distance and moves away horizontally at speed (half the launch speed). Gravity is . Positions (do not prove):
, .
Show that for the player to have a chance of hitting the target, must be less than of the maximum possible range of the projectile (to 2 significant figures).
How to recognise this question
Question type: vector motion
- The command and notation point to vector motion.
- The requested response is a show that.
How to handle it: differentiate position to get velocity and acceleration, then translate the angle condition into a dot product.
Watch out: Do not select a formula until its domain, interval, sign and units match the question.
Step-by-step answer
- Hit condition: some with equal positions: and .
- From the horizontal equation: (need ).
- From the vertical: (using ).
- Equate/substitute to get .
- Maximum free range of a projectile with speed is (at ).
- Thus .
- Maximise on : (using ).
- Positive root , giving .
- Hence must be less than about , i.e. less than of the maximum range.
- NESA: 4 marks; 3 for in terms of max range; 2 for max range; 1 for attempting flight time.
Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.
The paper records 4 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.
Try your hand at it again by answering these questions
Question 1
From the given and , show for a hit.
Question 2
A projectile launched at speed has maximum range . Prove it.
Question 3
Show that critical points of satisfy .