Stepwise
Start practisingPractise

Momentum, projectiles, work & energy · Lesson 2 of 2

Projectiles

A body projected at an angle: split the velocity into horizontal and vertical parts to find the time of flight, greatest height, range and position at any time.

14 minYou should already know: Kinematics & dynamics
  1. 1
  2. 2

A body projected at speed uu and angle θ\theta above the horizontal moves in two independent ways at once:

  • across, at a steady ucos⁡θu\cos\theta (nothing pushes it sideways);
  • up and down, starting at usin⁡θu\sin\theta and slowed by gg, as in motion under gravity.
θuu cos θu sin θHR
The path of a projectilex = (u cos θ)t; y = (u sin θ)t − ½gt²

Time, height and range

From level ground:

  • It lands when y=0y = 0: t(usin⁡θ−12gt)=0t\left(u\sin\theta - \frac12gt\right) = 0, so the time of flight is T=2usin⁡θgT = \dfrac{2u\sin\theta}{g}.
  • It is highest when the upward velocity is 00: 0=u2sin⁡2θ−2gH0 = u^2\sin^2\theta - 2gH, so H=u2sin⁡2θ2g{H = \dfrac{u^2\sin^2\theta}{2g}}.
  • The range is the distance across in time TT: R=ucos⁡θ×2usin⁡θg=u2sin⁡2θgR = u\cos\theta \times \dfrac{2u\sin\theta}{g} = \dfrac{u^2\sin 2\theta}{g}.

For a given speed the range is greatest at θ=45∘{\theta = 45^\circ}, where sin⁡2θ=1{\sin 2\theta = 1}.

A projectileSet the speed and the angle
HR
3.21 stime of flight12.91 mgreatest height61.55 mrange
Across: 19.15 m/s all the time. Up: 16.07 m/s, slowed by g. T = 2(25) sin 40° ÷ 10 = 3.21 s; H = 25² sin² 40° ÷ 20 = 12.91 m; R = 25² sin 80° ÷ 10 = 61.55 m. Angles 40° and 50° give the same range.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q14 (a)

A particle is projected at an angle of 42∘42^\circ with a velocity of 68 m s−168\text{ m s}^{-1}. Calculate, correct to one decimal place, the: (i) greatest height travelled; (ii) time of flight; (iii) horizontal range travelled. [g=10 m s−2][g = 10\text{ m s}^{-2}] Enter the three values.

  1. Greatest height

    • H=682sin⁡242∘2×10=4624×0.447720≈103.5{H = \frac{68^2\sin^2 42^\circ}{2 \times 10} = \frac{4624 \times 0.4477}{20} \approx 103.5} m.

    Think first. u = 68, θ = 42°, g = 10.

  2. Time of flight

    • T=2×68×sin⁡42∘10=136×0.669110≈9.1{T = \frac{2 \times 68 \times \sin 42^\circ}{10} = \frac{136 \times 0.6691}{10} \approx 9.1} s.

  3. Range

    • R=682sin⁡84∘10=4624×0.994510≈459.9{R = \frac{68^2\sin 84^\circ}{10} = \frac{4624 \times 0.9945}{10} \approx 459.9} m.

    Think first. sin 2θ = sin 84°.

Position at a given time or distance

To find where the body is, use x=(ucos⁡θ)tx = (u\cos\theta)t to get the time from the distance across, then put that time into y=(usin⁡θ)t−12gt2y = (u\sin\theta)t - \frac12gt^2.

Your turn

WAEC 2020 · Paper 2 · Q15 (b)

  1. (b)

    A particle is projected from a point PP with an initial velocity of 60 m s−160\text{ m s}^{-1} at an angle of 60∘60^\circ to the horizontal. Find its vertical displacement when its horizontal displacement is 7575 metres. [Take g=10 m s−2][\text{Take } g = 10\text{ m s}^{-2}]

Try it on a graph

Velocity–time graph: the area under it is the distance.

Worked solution (try it first)

(b)

  1. Horizontally: x=(60cos⁡60∘)t=30tx = (60\cos60^\circ)t = 30t, so t=2.5 st = 2.5\text{ s} when x=75x = 75.
  2. Vertically: y=(60sin⁡60∘)(2.5)−5(2.5)2y = (60\sin60^\circ)(2.5) - 5(2.5)^2
    =753−31.25= 75\sqrt3 - 31.25
    ≈98.65 m\approx 98.65\text{ m}.

Report a problem with this question