Hard A-Level Projectiles Questions

Challenging, exam-style A-Level Projectiles questions with worked solutions. Stretch yourself on the hardest horizontal-projection, speed, range, angled-projection problems.

horizontal-projectionspeedrangeangled-projectiontime-of-flightmaximum-height
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
If a projectile's horizontal range is RR when launched at speed uu from ground level, then sin⁡2α\sin 2\alpha equals
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Worked solution

  1. Start from the range formula

    R=u2sin⁡2αgR=\dfrac{u^2\sin 2\alpha}{g}

    Standard result for level ground.

  2. Rearrange

    sin⁡2α=Rgu2\sin 2\alpha=\dfrac{Rg}{u^2}

    Make sin⁡2α\sin 2\alpha the subject.

  3. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  4. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  5. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  6. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  7. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  8. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  9. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  10. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  11. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  12. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  13. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  14. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  15. Select the correct formula

    Rgu2\dfrac{Rg}{u^2}

    This inversion is used to find the launch angle for a given range.

Answer
Rgu2\dfrac{Rg}{u^2}
Question 2
8 markschallenging
A particle is projected at 52 m s−152\,\text{m s}^{-1} at an angle of 35∘35^{\circ} above the horizontal. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the greatest height reached. Give your answer to 3 significant figures.
Show worked solution

Worked solution

  1. Resolve the initial velocity

    uy=29.8 m s−1u_y=29.8\,\text{m s}^{-1}

    Vertical component determines the rise.

  2. At the top, vertical speed is zero

    vy=0 m s−1v_y=0\,\text{m s}^{-1}

    The projectile momentarily stops rising.

  3. Use vy2=uy2+2asv_y^2=u_y^2+2as with a=−9.8a=-9.8

    0=uy2−19.6H0=u_y^2-19.6H

    Relate initial and final vertical speeds to the height.

  4. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  5. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  6. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  7. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  8. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  9. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  10. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  11. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  12. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  13. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  14. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  15. State the greatest height

    H=45.4 mH=45.4\,\text{m}

    Maximum height above the point of projection.

Answer
H=45.4 mH=45.4\,\text{m}
Question 3
8 markschallenging
Neglecting air resistance, the speed of a projectile launched at angle α\alpha above the horizontal is least
Show worked solution

Worked solution

  1. Speed formula

    v=vx2+vy2v=\sqrt{v_x^2+v_y^2}

    Magnitude from the components.

  2. At the apex

    vy=0,  vx=ucos⁡αv_y=0,\; v_x=u\cos\alpha

    Vertical component vanishes; horizontal remains.

  3. Compare with launch

    vlaunch=u>ucos⁡αv_{\text{launch}}=u>u\cos\alpha

    Launch speed exceeds the speed at the top.

  4. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  5. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  6. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  7. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  8. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  9. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  10. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  11. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  12. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  13. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  14. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  15. Select the correct statement

    at the highest point of the path

    Speed is minimum when only the horizontal component remains.

Answer
at the highest point of the path
Question 4
8 markschallenging
For horizontal projection from a height hh at speed uu, the horizontal range is
Show worked solution

Worked solution

  1. Time of flight

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

    From vertical motion with uy=0u_y=0.

  2. Horizontal distance

    R=utR=ut

    Constant horizontal speed.

  3. Note the vertical motion

    uy=0,  a=9.8 m s−2u_y=0,\; a=9.8\,\text{m s}^{-2}

    There is no initial vertical speed for horizontal projection.

  4. Take downward as positive for the vertical part

    down=+\text{down}=+

    This keeps the acceleration positive.

  5. Separate horizontal and vertical motion

    x=ut,  y=12gt2x=ut,\; y=\tfrac12 gt^2

    Horizontal speed stays constant; vertical motion is suvat.

  6. Recall that horizontal speed is unchanged

    vx=uv_x=u

    Gravity does not affect the horizontal component.

  7. Check units

    t in s, R in mt\ \text{in s},\ R\ \text{in m}

    Times in seconds, distances in metres.

  8. Note the vertical motion

    uy=0,  a=9.8 m s−2u_y=0,\; a=9.8\,\text{m s}^{-2}

    There is no initial vertical speed for horizontal projection.

  9. Take downward as positive for the vertical part

    down=+\text{down}=+

    This keeps the acceleration positive.

  10. Separate horizontal and vertical motion

    x=ut,  y=12gt2x=ut,\; y=\tfrac12 gt^2

    Horizontal speed stays constant; vertical motion is suvat.

  11. Recall that horizontal speed is unchanged

    vx=uv_x=u

    Gravity does not affect the horizontal component.

  12. Check units

    t in s, R in mt\ \text{in s},\ R\ \text{in m}

    Times in seconds, distances in metres.

  13. Note the vertical motion

    uy=0,  a=9.8 m s−2u_y=0,\; a=9.8\,\text{m s}^{-2}

    There is no initial vertical speed for horizontal projection.

  14. Take downward as positive for the vertical part

    down=+\text{down}=+

    This keeps the acceleration positive.

  15. Select the correct formula

    R=u2hgR = u\sqrt{\dfrac{2h}{g}}

    Substituting for tt gives R=u2h/gR=u\sqrt{2h/g}.

Answer
R=u2hgR = u\sqrt{\dfrac{2h}{g}}
Question 5
8 markschallenging
Two projectiles are fired from ground level with the same speed but different angles. The one launched at 75∘75^{\circ} compared with the one at 15∘15^{\circ} will have
Show worked solution

Worked solution

  1. Compare the ranges

    R=u2sin⁡2αgR=\dfrac{u^2\sin 2\alpha}{g}

    sin⁡30∘=sin⁡150∘\sin 30^{\circ}=\sin 150^{\circ}, so ranges match.

  2. Compare maximum heights

    H=u2sin⁡2α2gH=\dfrac{u^2\sin^2\alpha}{2g}

    sin⁡75∘>sin⁡15∘\sin 75^{\circ}>\sin 15^{\circ}, so the steep launch goes higher.

  3. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  4. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  5. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  6. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  7. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  8. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  9. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  10. Vertical acceleration is gravity

    ay=−9.8 m s−2a_y=-9.8\,\text{m s}^{-2}

    Taking upward as positive, gravity is negative.

  11. Use independence of components

    x and y analysed separatelyx\ \text{and}\ y\ \text{analysed separately}

    The motions in the two directions do not affect each other.

  12. State g=9.8 m s−2g=9.8\,\text{m s}^{-2}

    g=9.8g=9.8

    Use this value throughout unless told otherwise.

  13. Resolve the initial velocity

    ux=ucos⁡α,  uy=usin⁡αu_x=u\cos\alpha,\; u_y=u\sin\alpha

    Split the launch speed into horizontal and vertical parts.

  14. Horizontal acceleration is zero

    ax=0a_x=0

    No horizontal force, so uxu_x stays constant.

  15. Select the correct statement

    a greater maximum height but the same range

    Complementary angles share range but not maximum height.

Answer
a greater maximum height but the same range

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