A small ball is projected with speed 25 m s⁻¹ at an angle of 30° above the horizontal from a point O on horizontal ground. At time t s after projection the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [B1] x = 25 cos 30t Use horizontal motion
2. [B1] y = 25 sin 30t - gt^2 / 2 Use vertical motion
3. [M1] y = x tan 30 - g/(2(25cos30)^2) x^2 Eliminate t
4. [A1] y = x/√3 - 4x^2 / 375 or y = 0.577x - 0.0107x^2
5. [M1A1] dy/dx = 1/√3 - 8x / 375 or dy/dx = 0.577 - 0.0214x Differentiate the equation from part (i) to find the gradient
6. [M1] -tan 15 = 1/√3 - 8x / 375 or -tan 15 = 0.577 - 0.0214x Attempt to solve
7. [A1] x = 39.6 or x = 39.5
8. Alternative method for question 4(ii)
9. [M1] tan 15 = v_y / v_x = v_y / (12.5√3)
10. [A1] v_y = -12.5√3 tan 15 (= -5.8) downwards
11. [M1] -5.8 = 12.5 - 10t leading to t = 1.83 Vertical motion using v = u + at
12. [A1] X = 1.83 × (25√3 / 2) = 39.6