AshPhys

Understanding projectiles

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.14Sections 2.14

Goal: Solve problems for a projectile launched at an angle by resolving its initial velocity.

velocity

01The big picture

Resolve the launch velocity into ucos⁡θu\cos\theta horizontally (constant all flight) and usin⁡θu\sin\theta vertically (decelerated by gg on the way up). At the highest point the vertical velocity is zero, but the horizontal velocity is not.

Over level ground the flight is symmetrical: time up = time down. Without air resistance the range is greatest at 45°. Air resistance lowers the height and the range and makes the descent steeper.

Range [paraphrase]
The horizontal distance travelled before landing.
Fig. 1 · Launched at an angle
Launched at an angleA projectile is launched from level ground at angle theta with velocity u, resolved into dashed components u cos theta and u sin theta. Its parabolic path reaches maximum height H halfway, where the velocity is horizontal, and lands at range R.θuu sin θu cos θu cos θHR
At the top only ucos⁡θu\cos\theta is left: the vertical velocity is zero.

02Key equations

Components
ux=ucos⁡θ,uy=usin⁡θu_x = u\cos\theta,\quad u_y = u\sin\theta
Level ground
T=2uyg,R=uxTT = \frac{2u_y}{g},\quad R = u_x T
Max height
H=uy22gH = \frac{u_y^{2}}{2g}
Symbols and units
  • u, θu,\ \thetalaunch speed and anglem s−1, ∘\mathrm{m\,s^{-1}},\ ^\circ
  • TTtime of flights\mathrm{s}
  • H, RH,\ Rmax height, rangem\mathrm{m}

03Sample problems with solutions

Sample 1 · A kicked ball
A ball is kicked at 20 m s⁻¹ at 30° above level ground. Find its time of flight, range and maximum height.

  1. Components:

    ux=20cos⁡30∘=17.3 m s−1,uy=20sin⁡30∘=10.0 m s−1u_x = 20\cos 30^\circ = 17.3\ \mathrm{m\,s^{-1}}, \quad u_y = 20\sin 30^\circ = 10.0\ \mathrm{m\,s^{-1}}

  2. Vertical gives the time; horizontal the range:

    T=2×10.09.81=2.04 s,R=17.3×2.04=35.3 mT = \frac{2 \times 10.0}{9.81} = 2.04\ \mathrm{s}, \quad R = 17.3 \times 2.04 = 35.3\ \mathrm{m}

  3. At the top vy=0v_y = 0:

    H=10.022×9.81=5.10 mH = \frac{10.0^{2}}{2 \times 9.81} = 5.10\ \mathrm{m}
Answer: 2.0 s; 35 m; 5.1 m
Unit check: (m s−1)2m s−2=m ✓\frac{(\mathrm{m\,s^{-1}})^{2}}{\mathrm{m\,s^{-2}}} = \mathrm{m}\ \checkmark

04Practice problems

  1. P1
    What is the velocity of a projectile at its highest point?
    Final answerucos⁡θu\cos\theta, horizontal
    Answer: ucos⁡θu\cos\theta, horizontal
  2. P2
    Thrown at 25 m s⁻¹ at 60°: find the maximum height.
    Final answer23.9 m (24 m)
    Answer: 23.9 m (24 m)
  3. P3
    At 15 m s⁻¹ and 40°: find the time of flight.
    Final answer1.97 s (2.0 s)
    Answer: 1.97 s (2.0 s)
  4. P4
    Which launch angle gives the longest range?
    Final answer45° (no air resistance)
    Answer: 45° (no air resistance)

05Other useful information

Common mistakes

  • ✗ Wrong: Use uu in the vertical equations.

    ✓ Right: Vertical: usin⁡θu\sin\theta. Horizontal: ucos⁡θu\cos\theta.

Exam tips

  • Find TT from the vertical motion first; it links the two directions.

What the syllabus asks

  • 9702 topic 2: projectile motion without air resistance. Coursebook 2.14.

Explore: Practice questions for this lesson