Projectile Motion: Time of Flight, Maximum Height and Range

Further Mathematics · Why horizontal and vertical motion are independent, resolving the launch velocity, the formulas for time of flight, greatest height and range, the 45° angle for maximum range, and horizontal projection, with WAEC worked examples.

What is a projectile?

A projectile is a body moving under gravity alone after it has been launched: a kicked football, a thrown stone, a bullet. Its path is a curve called a parabola.

Projectile vocabulary: launch speed u at angle θ, maximum height H, range R and time of flight T
Projectile vocabulary: launch speed u at angle θ, maximum height H, range R and time of flight T
  • Angle of projection (θ): the angle between the initial velocity and the horizontal
  • Time of flight (T): the total time in the air
  • Maximum (greatest) height (H): the highest point reached
  • Range (R): the horizontal distance travelled when it lands

The big idea: horizontal and vertical are independent

Gravity acts only vertically. It never speeds up or slows down the horizontal motion. So we treat the two directions separately:

Horizontal (no acceleration):

  • ux = u cos θ, which stays constant
  • x = ux t

Vertical (acceleration g downwards):

  • uy = u sin θ
  • vy = uy − gt
  • y = uyt − ½gt2

The three projectile formulas

For launch speed u at angle θ, landing at the same height:

  • Time of flight: T = 2u sin θ / g
  • Maximum height: H = u2 sin2 θ / 2g
  • Range: R = u2 sin 2θ / g

Which angle gives the greatest range?

Range at different angles for the same launch speed: greatest at 45°, and complementary angles give the same range
Range at different angles for the same launch speed: greatest at 45°, and complementary angles give the same range
  • R = u2 sin 2θ / g is greatest when sin 2θ = 1, i.e. θ = 45°. Then Rmax = u2/g.
  • Complementary angles (such as 30° and 60°) give the same range, but the larger angle goes higher and stays in the air longer.

Horizontal projection

If a body is launched horizontally (θ = 0) from a height h, then uy = 0:

  • time to fall: h = ½gt2
  • horizontal distance: x = ut

Velocity at any time

At time t: vx = u cos θ and vy = u sin θ − gt. The speed is √(vx2 + vy2), at angle tan−1(vy/vx) to the horizontal.

Common mistakes

  1. Applying g to the horizontal motion: horizontally, a = 0.
  2. Using the range formula when the projectile lands at a different height from where it started: use the component equations instead.
  3. Saying the velocity is zero at the top: only the vertical part is zero.
  4. Forgetting that sin 2θ means sin(2θ), not 2 sin θ.

Practice

  1. A ball is launched at 12 m/s at 45°. Find its time of flight (g = 10 m/s2).
  2. For u = 25 m/s and θ = 30°, find T, H and R (g = 10 m/s2).
  3. For u = 40 m/s and θ = 60°, find T, H and R (g = 10 m/s2).
  4. Find the maximum possible range for a launch speed of 30 m/s (g = 10 m/s2).
  5. A stone is thrown horizontally at 8 m/s from a cliff 20 m high. How far from the foot of the cliff does it land? (g = 10 m/s2)

Answers: 1. T = 2 × 12 × 0.707/10 ≈ 1.70 s · 2. T = 2.5 s, H ≈ 7.81 m, R ≈ 54.1 m · 3. T ≈ 6.93 s, H = 60 m, R ≈ 138.6 m · 4. Rmax = u2/g = 90 m · 5. 20 = 5t2, so t = 2 s and x = 16 m

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