A-level · Kinematics

Vertical motion under gravityExam-style questions

Why an object moving freely under gravity is just a constant-acceleration problem with a = ±g, how to choose a positive direction and translate "thrown up", "returns", "hits the ground" into suvat values, and why the up-and-down trip makes distance and displacement differ.

13 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4
  1. 1
    A small rock is thrown vertically upwards with speed 12.6 m s−112.6\ \text{m s}^{-1} from a point AA at the top of a vertical quarry face. The rock later lands at the point BB on the horizontal floor of the quarry, where BB is 2828 m vertically below AA. The rock does not bounce when it lands.

    The rock is modelled as a particle moving freely under gravity.
    (a) Find the speed of the rock at the instant it reaches BB.

    (3 marks)
    (b) Find the time taken for the rock to travel from AA to BB.

    (3 marks)
    (c) Find the length of time for which the rock is more than 66 m above the level of AA.

    (4 marks)
    (d) Sketch a speed-time graph for the motion of the rock from the instant it is thrown until the instant it lands at BB.

    (No further calculations are required.)

    (2 marks)
    Level 4 · 12 marksTry it
  2. 2
    A small stone is dropped from a bridge and falls vertically towards the river below.

    Before it reaches the river, the stone passes two fixed points, PP and QQ, with QQ below PP

    The speed of the stone is u m s−1u\ \text{m s}^{-1} as it passes PP and 5u m s−15u\ \text{m s}^{-1} as it passes QQ

    The only force acting on the stone is its weight, and the acceleration due to gravity, g m s−2g\ \text{m s}^{-2}, is constant.
    (a) Show that the stone takes 4ug\dfrac{4u}{g} seconds to fall from PP to QQ

    (2 marks)
    (b) The height of PP above the surface of the river is hh metres.

    Show that

    h>12u2gh > \frac{12u^2}{g}

    Fully justify your answer.

    (3 marks)
    Level 4 · 5 marksTry it
  3. 3
    In this question use g=9.8 m s−2g = 9.8\ \text{m s}^{-2}

    A fountain in a park has a nozzle at ground level. Water leaves the nozzle with a speed of 8.4 m s−18.4\ \text{m s}^{-1}

    A sign beside the fountain states that the water rises to a height of 3.63.6 metres above the ground.
    (a) Assume that the water leaves the nozzle vertically upwards.
    (a) (i) Verify the statement on the sign.

    (ii) State two modelling assumptions on which your answer to part (a)(i) depends.

    (4 marks)
    (b) In fact the water leaves the nozzle at an angle θ\theta to the vertical, where 0∘≤θ<20∘0^\circ \le \theta < 20^\circ

    The greatest height above the ground, hh metres, that the water reaches satisfies

    k<h≤3.6k < h \le 3.6

    Find the value of kk

    (4 marks)
    Level 4 · 8 marksTry it