A-LevelMathematicsMechanics9709/52 · Feb/Mar 2019 · Q36 Marks
A small ball is projected from a point O on horizontal ground. At time ts after projection the horizontal and vertically upwards displacements of the ball from O are xm and ym respectively, where x = 4t and y = 6t – 5t².
- (i)Find the equation of the trajectory of the ball.[2]
- (ii)Hence or otherwise calculate the angle of projection of the ball and its initial speed.[4]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2019). The mark or mark type is shown in brackets.
- [M1] Use horizontal and vertical motion and attempt to eliminate t
- y [= 6x/4 – 5 (x / 4)² ] = 1.5x –5 x²/16 or 1.5x – 0.3125 x²
- [A1]
- [M1] Use the trajectory equation from the formula sheet
- tan0 = 1.5
- [A1]
- 0 = 56.3°
- [M1] Again use the trajectory equation
- V²cos² 56.3 =16
- [A1]
- V = 7.21 m s-1
- OR
- [M1] Initial horizontal and vertical velocities
- Vcose = 4 and Vsin0 =6
- [M1] Use Pythagoras's theorem or trigonometry of a right angled triangle
- V²cos²0 + V²sin²0=4² + 6² OR tan0 = 6/4
- [A1]
- V = 7.21 m s-1
- [A1]
- 0 = 56.3°
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