A-Level Projectiles Practice Questions

Free A-Level Projectiles practice questions with full step-by-step worked solutions. Covers horizontal-projection, time-of-flight, range, formula-choice. Practise exam-style problems and check your method.

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.
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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
The path of a projectile launched at an angle above the horizontal is symmetric about
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Worked solution

  1. At the apex

    vy=0v_y=0

    Vertical speed is zero at the top.

  2. Equal ascent and descent times

    tup=tdownt_{\text{up}}=t_{\text{down}}

    For level ground, the motion up and down takes equal time.

  3. Select the correct statement

    the vertical line through the highest point

    The path is a symmetric parabola about its vertical axis through the apex.

Answer
the vertical line through the highest point
Question 3
3 marksintermediate
Complementary launch angles (e.g. 30∘30^{\circ} and 60∘60^{\circ}) with the same speed give the same horizontal range because
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Worked solution

  1. Write the range formula

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

    Range depends on the double angle.

  2. Compare 30∘30^{\circ} and 60∘60^{\circ}

    sin⁡60∘=sin⁡120∘\sin 60^{\circ}=\sin 120^{\circ}

    Supplementary double angles have equal sine.

  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. Select the correct statement

    sin⁡2α\sin 2\alpha is the same for both angles

    Equal sin⁡2α\sin 2\alpha means equal range for the same speed.

Answer
sin⁡2α\sin 2\alpha is the same for both angles
Question 4
5 markshard
In the trajectory equation y=xtan⁡α−gx22u2cos⁡2αy=x\tan\alpha-\dfrac{gx^2}{2u^2\cos^2\alpha}, the term −gx22u2cos⁡2α-\dfrac{gx^2}{2u^2\cos^2\alpha} represents
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Worked solution

  1. Identify the structure

    y=xtan⁡α⏟linear−gx22u2cos⁡2α⏟quadraticy=\underbrace{x\tan\alpha}_{\text{linear}}-\underbrace{\dfrac{gx^2}{2u^2\cos^2\alpha}}_{\text{quadratic}}

    Sum of a linear and a quadratic term in xx.

  2. Effect of the quadratic term

    negative for g,x,u>0\text{negative for }g,x,u>0

    It bends the path downward.

  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. Select the correct statement

    the downward curvature due to gravity

    Gravity produces the negative quadratic correction.

Answer
the downward curvature due to gravity
Question 5
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}

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