Projectiles Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Projectiles questions. See exactly how to solve problems on horizontal-projection, time-of-flight, range, formula-choice.

horizontal-projectiontime-of-flightrangeformula-choicetrajectoryconcept
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle is projected horizontally at 10 m s−110\,\text{m s}^{-1} from a point 20 m20\,\text{m} above level ground. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the time taken to reach the ground. Give your answer to 3 significant figures.

Worked solution

  1. List the known quantities

    u=10 m s−1,  h=20 mu=10\,\text{m s}^{-1},\; h=20\,\text{m}

    Horizontal speed and launch height are given.

  2. Use t=2hgt=\sqrt{\dfrac{2h}{g}}

    t=2×209.8t=\sqrt{\dfrac{2\times 20}{9.8}}

    Vertical fall from rest gives the time of flight.

  3. State the time of flight

    t=2.02 st=2.02\,\text{s}

    This is the time until the projectile lands.

Answer
t=2.02 st=2.02\,\text{s}
Question 2
2 markseasy
A ball is projected horizontally at 15 m s−115\,\text{m s}^{-1} from the top of a cliff 45 m45\,\text{m} high. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find how far horizontally it lands from the base of the cliff. Give your answer to 3 significant figures.

Worked solution

  1. Find the time of flight

    t=3.03 st=3.03\,\text{s}

    Use t=2h/gt=\sqrt{2h/g} from the vertical motion.

  2. Use R=utR=ut

    R=15×3.03R=15\times 3.03

    Horizontal speed is constant.

  3. State the horizontal range

    R=45.5 mR=45.5\,\text{m}

    This is the horizontal distance travelled.

Answer
R=45.5 mR=45.5\,\text{m}
Question 3
2 markseasy
A particle is projected horizontally at 8 m s−18\,\text{m s}^{-1} from a point 19.6 m19.6\,\text{m} above level ground. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the time taken to reach the ground. Give your answer to 3 significant figures.

Worked solution

  1. List the known quantities

    u=8 m s−1,  h=19.6 mu=8\,\text{m s}^{-1},\; h=19.6\,\text{m}

    Horizontal speed and launch height are given.

  2. Use t=2hgt=\sqrt{\dfrac{2h}{g}}

    t=2×19.69.8t=\sqrt{\dfrac{2\times 19.6}{9.8}}

    Vertical fall from rest gives the time of flight.

  3. State the time of flight

    t=2 st=2\,\text{s}

    This is the time until the projectile lands.

Answer
t=2 st=2\,\text{s}
Question 4
2 markseasy
A particle is projected horizontally from a height. Which formula gives the time tt until it reaches level ground?

Worked solution

  1. Identify the vertical motion

    uy=0,  s=hu_y=0,\; s=h

    The vertical part starts from rest.

  2. Choose a suvat equation with uy=0u_y=0

    h=12gt2h=\tfrac12 gt^2

    Displacement under constant acceleration from rest.

  3. Select the correct formula

    t=2hgt = \sqrt{\dfrac{2h}{g}}

    This gives the time of flight for horizontal projection.

Answer
t=2hgt = \sqrt{\dfrac{2h}{g}}
Question 5
2 markseasy
A ball is projected horizontally at 12 m s−112\,\text{m s}^{-1} from the top of a cliff 30 m30\,\text{m} high. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find how far horizontally it lands from the base of the cliff. Give your answer to 3 significant figures.

Worked solution

  1. Find the time of flight

    t=2.47 st=2.47\,\text{s}

    Use t=2h/gt=\sqrt{2h/g} from the vertical motion.

  2. Use R=utR=ut

    R=12×2.47R=12\times 2.47

    Horizontal speed is constant.

  3. State the horizontal range

    R=29.7 mR=29.7\,\text{m}

    This is the horizontal distance travelled.

Answer
R=29.7 mR=29.7\,\text{m}

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