Vertical projection (thrown up)

Thrown up at u: time up = u/g, max height = u²/2g, and the return speed equals u.

Vertical projection (thrown up)0 min · Free lecture

In this lesson

  • Thrown up at u: time up = u/g, max height = u²/2g, and the return speed equals u.

The journey up is the journey down in reverse — same times, same speeds.

With up as positive, a = −g. Time to the top: t = u/g (v = 0 there). Maximum height: H = u²/2g. Total flight time: 2u/g.

Symmetry: at the same height on the way up and down, speeds are equal and opposite; the time from the top down to any height equals the time up from that height to the top.

T=2ug,H=u22gT = \frac{2u}{g},\quad H = \frac{u^2}{2g}

Flight time and max height for a vertical throw

Worked example

The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.

  1. Area under the a–t graph = change in velocity.
  2. Area = 2 m/s² × 4 s = 8 m/s (the last 2 s add nothing).

Answer: 8

Transcript (0 min)
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Vertical projection (thrown up) — FemtoLearn.
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Thrown up at u: time up = u/g, max height = u²/2g, and the return speed equals u.
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Practice

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

    The acceleration of a body starting from rest is a = 2 m/s² for 4 s, then zero for 2 s. Find its velocity at t = 6 s.

  2. Q2 · MCQ · difficulty 2/5

    A particle moves with constant velocity. Which a–t graph describes it?

    • A horizontal line at a = 0
    • A horizontal line at a = 2 m/s²
    • A straight line through the origin
    • A parabola
  3. Q3 · Numerical · difficulty 2/5

    A body moves with v = 10 m/s for 3 s, then v = 5 m/s for 2 s in the same direction. Find the total distance.

  4. Q4 · MCQ · difficulty 2/5

    The slope of a velocity–time graph gives:

    • Displacement
    • Acceleration
    • Speed
    • Distance