Speed and direction at any time

v = √(uₓ² + v_y²) with v_y = u sinθ − gt; direction from tanφ = v_y/uₓ.

Speed and direction at any time0 min · Free lecture

In this lesson

  • v = √(uₓ² + v_y²) with v_y = u sinθ − gt; direction from tanφ = v_y/uₓ.

The projectile's speed changes every instant, but its horizontal component never does.

At time t: vₓ = u cosθ (constant), v_y = u sinθ − gt. Speed: v = √(vₓ² + v_y²); angle to the horizontal: tanφ = v_y/vₓ.

The velocity is tangent to the trajectory — matching the trajectory slope is a good cross-check.

v=(ucos⁡θ)2+(usin⁡θ−gt)2v = \sqrt{(u\cos\theta)^2 + (u\sin\theta - gt)^2}

Speed at time t

Worked example

For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

  1. R = v0² sin 2θ / g
  2. R is maximum when sin 2θ = 1, i.e. 2θ = 90°, so θ = 45°

Answer: 45°

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Speed and direction at any time — FemtoLearn.
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v = √(uₓ² + v_y²) with v_y = u sinθ − gt; direction from tanφ = v_y/uₓ.
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Practice

Free · 4 questions with full solutions
  1. Q1 · Numerical · difficulty 2/5

    A projectile is launched with speed 15 m/s at an angle 70° above the horizontal. Taking g = 10 m/s², find its horizontal range.

    R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g}
  2. Q2 · Numerical · difficulty 2/5

    A projectile is launched with speed 15 m/s at an angle 25°. Taking g = 10 m/s², find its maximum height above the launch point.

    H=v02sin⁡2θ2gH = \frac{v_0^2 \sin^2\theta}{2g}
  3. Q3 · Numerical · difficulty 2/5

    A projectile is launched with speed 25 m/s at an angle 15°. Taking g = 9.8 m/s², find its total time of flight.

  4. Q4 · MCQ · difficulty 1/5

    For a fixed launch speed, the horizontal range of a projectile is maximum when the launch angle is:

    • 30°
    • 45°
    • 60°
    • 90°