A-LevelMathematicsMechanics9709/52 · Feb/Mar 2018 · Q49 Marks

A particle P is projected from a point O on horizontal ground. At the instant t s after projection, the horizontal and vertically upwards displacements of P from O are x m and y m respectively. The equation of the trajectory of P is y = 3x – 0.05x².

(i)Find the angle of projection and the initial speed of P.[3]
(ii)Find the coordinates of P at the instant when OP makes an angle of 45° with the horizontal.[2]
(iii)For the instant when P is at its greatest height above the ground, calculate this height and the corresponding value of t.[4]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.

[B1] (tanθ = 3) θ = 71.6° Use the formula sheet for the trajectory equation
[M1] 0.05 = g / (2V²cos²71.6)
[A1] V = 10√10 = 31.6 m s⁻¹
[M1] Use y = x
[A1] x = 40 and y = 40
[M1] dy/dx = 3 -0.1x = 0 Use the fact that the gradient is zero at the highest point
[A1] x = 30, y = (3 × 30 – 0.05 × 30²) = 45
[M1] 30 = (31.6cos71.6)t Use horizontal motion
[A1] t = 3.01 t=3 if exact arithmetic used

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