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Lesson 7 · The height of a throw
The parabola you can throw
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The parabola you can throw

Throw a ball straight up at v0v_0 — v-nought, the launch speed — and its height follows h=v0t−12gt2h = v_0 t - \tfrac{1}{2} g t^2. Read aloud: h equals v-nought t minus one-half g t-squared, with g=9.8 m/s2g = 9.8\ \mathrm{m/s^2}, gravity's tax rate. Look hard at that formula: it IS y=ax2+bx+cy = ax^2 + bx + c wearing physics clothes — a = −4.9, b = v₀, c = 0. Everything this chapter taught now works on the sky: the apex is the vertex, the landing is a root.

Two shortcuts fall straight out: the peak is H=v022gH = \dfrac{v_0^2}{2g} and the full flight lasts T=2v0gT = \dfrac{2 v_0}{g} — up takes v0/gv_0/g, and the fall politely takes the same. Run the units as a guide, never a confession: (m/s)⋅s=m\mathrm{(m/s)\cdot s = m} and (m/s2)⋅s2=m\mathrm{(m/s^2)\cdot s^2 = m} — every term in a height equation must BE a height, or the equation is lying.

v0atd

d=v0t+12at2d = v_0 t + \tfrac{1}{2} a t^2

  • dd= Displacement (length)
  • v0v_0= Initial velocity (speed)
  • aa= Acceleration (acceleration)
  • tt= Time (time)
Displacement (Uniform Acceleration) solver →
θv0H

H=v02sin⁡2θ2gH = \frac{v_0^{2} \sin^{2}\theta}{2g}

  • HH= Maximum height (length)
  • v0v_0= Launch speed (speed)
  • θ\theta= Launch angle (angle)
Projectile Maximum Height solver →
θv0T

T=2v0sin⁡θgT = \frac{2 v_0 \sin\theta}{g}

  • TT= Time of flight (time)
  • v0v_0= Launch speed (speed)
  • θ\theta= Launch angle (angle)
Projectile Time of Flight solver →