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8.2MechanicsStretch

Vector Methods with Projectiles

Projectile motion can be handled neatly with vectors. The acceleration is the constant gravity vector −gj, and the suvat vector equations give position and velocity at any time.

24 min Further Kinematics
Edexcel A Level Maths: 8.2 Vector Methods with ProjectilesCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Use a = −gj for projectiles
  • Apply vector suvat to projectile motion
  • Find velocity and position vectors
  • Determine when conditions are met
1

Gravity as a vector

For a projectile, the acceleration is a=−gj\mathbf{a} = -g\mathbf{j} (taking up as positive j\mathbf{j}). The velocity and position then follow from the vector suvat equations.

a=−gj,v=u−gt j\mathbf{a} = -g\mathbf{j}, \quad \mathbf{v} = \mathbf{u} - g t\,\mathbf{j}
Gravity acts in the −j direction.
Example 1A projectile has u=7i+10j\mathbf{u} = 7\mathbf{i} + 10\mathbf{j}. Find v\mathbf{v} at t=1t = 1 (g = 9.8).
1v=7i+10j−9.8j\mathbf{v} = 7\mathbf{i} + 10\mathbf{j} - 9.8\mathbf{j}.
2=7i+0.2j= 7\mathbf{i} + 0.2\mathbf{j}.
Answer
7i+0.2j7\mathbf{i} + 0.2\mathbf{j}

Tip — The i component of velocity stays constant; only the j component changes (by −gt).

Formula recap

a=−gj\mathbf{a} = -g\mathbf{j}
Gravity vector.
r=ut−12gt2j\mathbf{r} = \mathbf{u}t - \tfrac12 g t^2\mathbf{j}
Position vector.

Common mistakes to avoid

Applying gravity to the i component.
Gravity is in the −j direction only.
Forgetting the sign of g.
With up positive, the acceleration is −gj.

Key takeaways

  • Projectile acceleration is a = −gj.
  • Use vector suvat for velocity and position.
  • The i (horizontal) component is unchanged.

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

  1. For a projectile, the acceleration vector is:

    • −gj-g\mathbf{j}
    • gig\mathbf{i}
    • −gi-g\mathbf{i}
    • gjg\mathbf{j}
    Show answer and explanation

    Answer: −gj-g\mathbf{j}

    Gravity downward.

  2. The horizontal (i) component of velocity:

    • stays constant
    • decreases
    • increases
    • reverses
    Show answer and explanation

    Answer: stays constant

    No horizontal acceleration.

  3. With u=7i+10j\mathbf{u}=7\mathbf{i}+10\mathbf{j}, v\mathbf{v} at t=1 is (g=9.8):

    • 7i+0.2j7\mathbf{i}+0.2\mathbf{j}
    • 7i+10j7\mathbf{i}+10\mathbf{j}
    • 0.2i+7j0.2\mathbf{i}+7\mathbf{j}
    • 7i−9.8j7\mathbf{i}-9.8\mathbf{j}
    Show answer and explanation

    Answer: 7i+0.2j7\mathbf{i}+0.2\mathbf{j}

    10 − 9.8 = 0.2.

  4. The position vector is:

    • ut−12gt2j\mathbf{u}t - \tfrac12 g t^2\mathbf{j}
    • u−gtj\mathbf{u} - gt\mathbf{j}
    • 12gt2j\tfrac12 g t^2\mathbf{j}
    • ut\mathbf{u}t
    Show answer and explanation

    Answer: ut−12gt2j\mathbf{u}t - \tfrac12 g t^2\mathbf{j}

    Vector suvat.

  5. The vertical velocity is zero at the:

    • highest point
    • launch
    • landing
    • start only
    Show answer and explanation

    Answer: highest point

    Top of the path.

Next step

Keep practising

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