AcademyMotion in Space
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Projectile Models

Level 1 - Physics topic page in Motion in Space.

Principle

Projectile motion uses one shared time variable with horizontal acceleration zero and vertical acceleration downward.

The model applies after launch when air resistance is ignored and gravity is the only force.

Notation

\(x,y\)
horizontal and vertical positions
\(\mathrm{m}\)
\(v_0\)
launch speed
\(\mathrm{m\,s^{-1}}\)
\(\theta\)
launch angle above horizontal
rad or deg
\(g\)
gravitational field strength
\(\mathrm{m\,s^{-2}}\)
\(R\)
level-ground range
\(\mathrm{m}\)

Method

Derivation 1: Resolve the launch velocity

The initial velocity is a vector. The horizontal component uses the adjacent projection; the vertical component uses the opposite projection.

Horizontal component
\[v_{0x}=v_0\cos\theta\]
Vertical component
\[v_{0y}=v_0\sin\theta\]
Initial velocity vector
\[\vec v_0=v_0\cos\theta\,\hat{\imath}+v_0\sin\theta\,\hat{\jmath}\]

Derivation 2: Separate the two component models

Gravity gives vertical acceleration only. The horizontal equation and vertical equation are linked by the same time variable.

Horizontal acceleration
\[a_x=0\Rightarrow v_x=v_0\cos\theta\]
Horizontal position
\[x=x_0+v_0\cos\theta\,t\]
Vertical position
\[y=y_0+v_0\sin\theta\,t-\frac12gt^2\]
Both component equations use the same \(t\).

Derivation 3: Derive the level-ground range

The range formula is not a general projectile formula. It assumes the landing height equals the launch height.

Set level-ground landing
\[0=v_0\sin\theta\,t-\frac12gt^2\]
Nonzero flight time
\[t_f=\frac{2v_0\sin\theta}{g}\]
Insert into horizontal motion
\[R=v_0\cos\theta\,t_f=\frac{v_0^2\sin2\theta}{g}\]
The last step uses \(2\sin\theta\cos\theta=\sin2\theta\).

The sketch shows the model separation: horizontal motion is uniform, while vertical motion curves because acceleration is downward.

0246810012345xytrajectoryv0
The path curves because only the vertical component accelerates.

Rules

These are the compact results from the derivations above.

Horizontal position
\[x=x_0+v_0\cos\theta\,t\]
Vertical position
\[y=y_0+v_0\sin\theta\,t-\frac12gt^2\]
Horizontal velocity
\[v_x=v_0\cos\theta\]
Vertical velocity
\[v_y=v_0\sin\theta-gt\]
Level-ground range
\[R=\frac{v_0^2\sin2\theta}{g}\]

Examples

Question
A projectile is launched at
\[20\,\mathrm{m\,s^{-1}}\]
and
\[30^\circ\]
from level ground. Find its flight time.
Answer
First resolve the vertical component:
\[v_{0y}=20\sin30^\circ=10\,\mathrm{m\,s^{-1}}\]
For level ground,
\[0=v_{0y}t-\frac12gt^2\]
The nonzero solution is
\[t=\frac{2v_{0y}}{g}=\frac{20}{9.8}=2.04\,\mathrm{s}\]

Checks

  • Horizontal and vertical equations share the same time.
  • Horizontal acceleration is zero only when air resistance is ignored.
  • Vertical acceleration is downward.
  • The range formula requires launch and landing at the same height.

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Acceleration Vectors

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Circular Motion Kinematics