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Projectile motionPhysicsHigh schoolabout 1:23

A projectile is launched at 15 m/s at 0° from a height of 20 m. Find its time of flight, range and maximum height: animated lesson

A narrated, step-by-step projectile motion video lesson with captions. Final answer: Time of flight ≈ 2.0192751 s; range ≈ 30.289127 m; maximum height ≈ 20 m; impact speed ≈ 24.847535 m/s.

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (0.6 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    A projectile is launched at 15 m/s at 0° from a height of 20 m. Find its time of flight, range and maximum height.
  2. 2. Steps, part 1
    1. vx=v0cos⁡θ=15cos⁡0∘=15,vy=v0sin⁡θ=0 m/sv_x = v_0\cos\theta = 15\cos 0^\circ = 15,\quad v_y = v_0\sin\theta = 0\ \text{m/s}
      Split the launch velocity into components
    2. tpeak=vyg=0 st_{\text{peak}} = \frac{v_y}{g} = 0\ \text{s}
      Time to the highest point
    3. H=h0+vy22g=20 mH = h_0 + \frac{v_y^2}{2g} = 20\ \text{m}
      Maximum height
    4. h0+vyt−12gt2=0⇒t=vy+vy2+2gh0g=2.0192751 sh_0 + v_y t - \tfrac12g t^2 = 0 \Rightarrow t = \frac{v_y + \sqrt{v_y^2 + 2gh_0}}{g} = 2.0192751\ \text{s}
      Time of flight (land at height 0)
  3. 3. Steps, part 2
    1. R=vxt=15×2.0192751=30.289127 mR = v_x t = 15 \times 2.0192751 = 30.289127\ \text{m}
      Horizontal range
    2. vy(T)=vy−gT=−19.809089,∣v∣=vx2+vy(T)2=24.847535 m/s at 52.866∘ below horizontalv_y(T) = v_y - gT = -19.809089,\quad |v| = \sqrt{v_x^2 + v_y(T)^2} = 24.847535\ \text{m/s at } 52.866^\circ \text{ below horizontal}
      Impact velocity
  4. 4. The answer
    T≈2.0192751 s, R≈30.289127 m, H≈20 mT \approx 2.0192751\ \text{s},\ R \approx 30.289127\ \text{m},\ H \approx 20\ \text{m}

    T = 2.0192751 s, R = 30.289127 m, H = 20 m

  5. 5. Check
    • Energy check: v² at impact = 617.4 and v₀² + 2gh₀ = 617.4 (equal).

Transcript

The problem. Here is the problem: A projectile is launched at 15 m/s at 0° from a height of 20 m. Find its time of flight, range and maximum height. Treat horizontal and vertical motion separately: constant velocity horizontally, constant downward acceleration g vertically, ignoring air resistance.

Steps, part 1. Split the launch velocity into components. Time to the highest point. The launch is level or downward, so the launch point is the highest point. Maximum height. Time of flight (land at height 0).

Steps, part 2. Horizontal range. Impact velocity.

The answer. So the answer is Time of flight approximately 2.0192751 s; range approximately 30.289127 m; maximum height approximately 20 m; impact speed approximately 24.847535 m/s. Without air resistance the horizontal velocity never changes, while the vertical velocity falls by g every second. The flight ends when the height returns to zero.

Check. Energy check: v squared at impact equals 617.4 and v₀ squared plus 2gh₀ equals 617.4 (equal).

About projectile motion

Time of flight, range and maximum height of a projectile.

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