A-LevelMathematicsMechanics9709/52 · Oct/Nov 2018 · Q48 Marks
A small object is projected horizontally with speed V m s⁻¹ from a point O above horizontal ground. At time t s after projection, the horizontal and vertically upwards displacements of the object from O are x m and y m respectively.
- (i)Express x and y in terms of t and hence show that the equation of the path of the object is y = -5x²/V².[3]
- (ii)The object passes through points with coordinates (a, −a) and (a², −16a), where a is a positive constant. Find the value of a.[3]
- (iii)Given that the object strikes the ground at the point where x = 5a, find the height of O above the ground.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [B1] x = Vt or y = -1/2 gt²
- [M1] Attempt to eliminate t
- [AG] y = -5x²/V²
- [B1] -a = -5 a²/V² or -16a = -5 (a²)²/V²
- [M1] Attempt to eliminate V
- [A1] a = 4
- [M1] Use the value of a found in part ii
- [A1] y = -5(5×4)²/(5 × 4) (= -100) [A1] Height = 100 m
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