Projectiles flashcards

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Projectile (in M2)
A particle that, once launched, moves freely under gravity alone in a vertical plane.
Angle of projection
The angle between the initial velocity and the horizontal.
Range (on level ground)
The horizontal distance from the point of projection to the point where the particle returns to the same level.
Time of flight
The time from projection until the particle lands (returns to the launch level, or hits the given surface).
Trajectory
The path followed by the projectile; with no air resistance it is a parabola.
Modelling assumptions for a projectile
Particle (no size or spin); no air resistance; g constant (9.8 m s⁻²); so the only force is the weight and the acceleration is g downwards.
Components of an initial velocity u at angle θ above the horizontal
Horizontal u cos θ; vertical u sin θ. (No — derive it)
Horizontal and vertical displacement at time t
x = (u cos θ)t; y = (u sin θ)t − ½gt². (No — derive it)
Constant-acceleration formulae used for the vertical motion
v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t, s = vt − ½at², with a = −9.8 (up positive). (No — must be known)
Time of flight on level ground
T = 2u sin θ / g. (No — derive it)
Range on level ground
R = u² sin 2θ / g; maximum u²/g at θ = 45°. (No — derive it)
Greatest height above the launch point
H = u² sin²θ / (2g). (No — derive it)
Equation of the trajectory
y = x tan θ − gx² / (2u² cos²θ) = x tan θ − (gx²/2u²)(1 + tan²θ). (No — derive it)
Vector form of position and velocity (from O)
r = ut − ½gt² j; v = u − gt j, with j vertically up. (No — derive it)
Shape of a projectile's path (no air resistance)
A parabola, symmetrical about the vertical through the highest point when launch and landing are at the same level.
Velocity components drawn at equal time intervals along the path
Horizontal arrows all equal; vertical arrows decrease by g × Δt each interval, are zero at the top, then grow downwards.
Paths at the same speed for 30° and 60°
Same range, but the 60° path is higher and takes longer.
Direction of the velocity on a path diagram
Along the tangent to the path at that point.
Why is the horizontal component of velocity constant?
The only force is the weight, which is vertical, so there is no horizontal acceleration.
Velocity at the greatest height
Vertical component zero; the particle still moves horizontally at u cos θ — its least speed.
Why do θ and 90° − θ give the same range?
sin 2θ = sin(180° − 2θ) = sin 2(90° − θ).
Why can there be two angles of projection through a given point?
Substituting the point into the trajectory gives a quadratic in tan θ; two positive roots give a low path and a high path.
What links the horizontal and vertical motions?
Time: the same t appears in both, so find t from one direction and use it in the other.
Does a heavier projectile follow a different path?
No (in the model): acceleration is g for every mass, so mass does not appear in the equations.
Projected horizontally at 5 m s⁻¹ from 19.6 m above level ground. Time and horizontal distance?
19.6 = 4.9t² → t = 2 s; distance = 5 × 2 = 10 m.
20 m s⁻¹ from level ground with sin θ = 0.6. Greatest height, time of flight, range?
Components 16 and 12 m s⁻¹. H = 12²/19.6 = 7.35 m; T = 24/9.8 = 2.45 s; R = 16 × 2.45 = 39.2 m.
Initial velocity (10i + 15j) m s⁻¹. Speed and direction after 2.5 s?
v = 10i − 9.5j; speed √(100 + 90.25) = 13.8 m s⁻¹ at 43.5° below the horizontal.
Greatest range on level ground at 14 m s⁻¹?
u²/g = 196/9.8 = 20 m, at 45°.
u = 20 m s⁻¹; the path passes through (20, 5). Quadratic in tan θ and the angles?
5 = 20 tan θ − 4.9(1 + tan²θ) → 4.9 tan²θ − 20 tan θ + 9.9 = 0 → θ = 30.0° or 74.1°.
From 25 m above the sea with vertical component 3 m s⁻¹ upwards. Time to reach the sea?
−25 = 3t − 4.9t² → 4.9t² − 3t − 25 = 0 → t = 2.59 s (reject the negative root).
Using R = u² sin 2θ / g when launched from a cliff
Wrong: it needs equal launch and landing levels. Find t from the vertical quadratic, then x = (u cos θ)t.
Sign of the displacement to a point below the launch
With up positive, it is negative (e.g. −30), and a = −9.8.
Giving a direction of motion
Give an angle and say above or below the horizontal (use the sign of the vertical component).
“Show that … is a quadratic in tan θ”
Write 1/cos²θ as sec²θ, then use sec²θ = 1 + tan²θ, and show every line.
Accuracy with g = 9.8
Give final answers to 2 or 3 s.f.; carry unrounded values between parts.
Two positive roots for t at a given height
Both can be valid: the particle passes that height on the way up and on the way down. Read which one the question wants.

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