A-LevelMathematicsMechanics9709/52 · Oct/Nov 2019 · Q46 Marks
A particle is projected from a point O on horizontal ground with speed Vms¯¹ at an angle of 60° above the horizontal. At the instant 3 s after projection the direction of motion of the particle is 30° below the horizontal.
- (i)Find V.[3]
- (ii)Calculate the distance of the particle from O at the instant 3 s after projection.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- [B1] Velocity component vertically = ± (Vsin60 – 3g)
- [M1] tan 30 = (3g - Vsin60) / (Vcos60)
- [A1] V=15√3 = 26(.0) m s⁻¹
- [B1FT] y = 26sin60 × 3 – g × 3² / 2
- [M1] D² = (26sin60 × 3 - g × 3²)² + (26cos60 × 3)²
- [A1] D=45(.0) m
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