Motion of a projectile

Further Maths · Further Mechanics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A projectile is modelled as a particle that moves under gravity only. Air resistance, spin and the size of the object are ignored, and g is taken as constant. Real objects therefore travel a little less far than the model predicts.

The idea to hold is that the horizontal and vertical motions are independent. Horizontally there is no force, so the horizontal velocity never changes. Vertically the acceleration is g downwards, so the constant acceleration equations apply. Time is the link between the two directions.

Removing t from the two equations gives the trajectory, a parabola connecting y and x. It is the best tool when a point on the path is given and the speed or angle is unknown.

Rules to remember

For initial speed V at angle θ above the horizontal: horizontal velocity = V cos θ (constant); vertical velocity = V sin θ − gt
Position: x = (V cos θ)t and y = (V sin θ)t − ½gt², with y measured upwards from the point of projection
Speed at any instant = √(v_x² + v_y²); direction to the horizontal from tan(angle) = v_y ÷ v_x
Greatest height = (V sin θ)²/(2g), reached when the vertical velocity is zero; the speed there is V cos θ, not zero
On level ground only: time of flight = (2V sin θ)/g and range = (V² sin 2θ)/g
Trajectory: y = x tan θ − gx²/(2V² cos² θ), which can be written y = x tan θ − gx²(1 + tan² θ)/(2V²)
Use g = 10 m s⁻² unless the question gives another value

Common mistake

Students use the range formula V² sin 2θ/g when the particle does not land at the level it started from, for example when it is thrown from a cliff. In that case find the time from the vertical motion first, then use x = (V cos θ)t.

Worked example

A particle is projected from level ground with speed 26 m s⁻¹ at an angle θ above the horizontal, where sin θ = 5/13. Take g = 10 m s⁻². Find its speed and direction of motion 1.7 s after projection.

cos θ = 12/13, so horizontal velocity = 26 × 12/13 = 24 m s⁻¹ (constant).
Vertical velocity = 26 × 5/13 − 10 × 1.7 = 10 − 17 = −7 m s⁻¹, so the particle is moving downwards.
Speed = √(24² + 7²) = √625 = 25 m s⁻¹.
tan(angle) = 7/24, so the angle is 16.3° below the horizontal.

Answer: 25 m s⁻¹ at 16.3° below the horizontal

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