Projectile given in i/j vector form - find the time or height at which the velocity is perpendicular to the initial launch velocity

IAL M2 · Vectors / projectiles · appears in 3 of 25 sittings, 2023-2024

A particle is projected with a given initial velocity in i/j vector form and moves freely under gravity. Find the time (or, from that, the height above the ground) at the instant when the particle's velocity is perpendicular to its initial velocity.

A projectile whose velocity must be perpendicular to its own velocity at another instant appears in 5 of 25 M2 sittings. In 2 of them (2020-October, 2021-June) the other instant is a later point or the landing, not the launch, so the same scalar-product step is applied to a velocity the student first has to find.

The method

4 marks at its longest. Each step below is a mark the examiner is looking for.

  1. Form the perpendicularity condition: the dot product of the initial velocity vector and the velocity vector at time t is zero; condone sign errors in setting this up (1 mark)
  2. Solve the resulting equation for the vertical velocity component, or directly for t, depending on what the question asks for (1 mark)
  3. Where a height is asked for: use suvat or an energy equation, dependent on the previous method mark, to convert the found velocity/time into a height above the ground (2 marks)

What changes between years

Everything below varies from sitting to sitting. The method above does not.

Where the marks go

Instructions the examiners are given, repeated across sittings.

Where it appeared

2023-June, 2023-October, 2024-January.

GOAL-15 Phase 3 recounted this over all 25 M2 sittings (it had been counted over 11). No sitting outside the three already on record asks for perpendicularity to the initial velocity.

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