Projectile motion: launched at an angle

Splitting the launch velocity into components

• When a projectile is launched at an angle $\theta$ above the horizontal with speed $v$, neither component is zero at the start.
• Resolve the launch velocity into two perpendicular components:

$$v_x = v\cos\theta \qquad v_y = v\sin\theta$$

• $v$ — launch speed (m s⁻¹)

• $\theta$ — launch angle above the horizontal

• $v_x$ — horizontal component, constant for the whole flight

• $v_y$ — initial vertical component, which then behaves like a ball thrown straight up

• Cosine goes with the horizontal, sine goes with the vertical — when the angle is measured from the horizontal.

What each component does during the flight

• Horizontal: no force, so $v_x$ never changes. Horizontal displacement is always $d_x = v_x t$.
• Vertical: starts at $v_y$ upward, is slowed by gravity, reaches zero at the top, then grows downward.
• The path is symmetric about the highest point when the launch and landing heights are the same:
  • time up = time down,
  • the speed at landing equals the launch speed,
  • the landing angle below the horizontal equals the launch angle above it.

The three quantities you are usually asked for

Maximum height — work in the vertical column with $v_{fy} = 0$ at the top:

$$v_{fy}^2 = v_y^2 + 2ad_y \;\Rightarrow\; d_y = \frac{v_y^2}{2g}$$

Time of flight — find the time to the top, then double it (equal launch and landing heights):

$$t_{\text{top}} = \frac{v_y}{g}, \qquad t_{\text{total}} = \frac{2v_y}{g}$$

Range — the horizontal distance covered in the whole flight:

$$d_x = v_x \, t_{\text{total}}$$

• If the projectile lands at a different height from its launch (off a cliff, into a hole), the symmetry shortcut fails. Solve the vertical motion properly with $d_y$ equal to the height difference, using signs consistently.

How the launch angle changes the range

• For a given launch speed on level ground:
  • the range is greatest at $45°$,
  • complementary angles give the same range — $30°$ and $60°$ land in the same place,
  • a steeper angle gives more height and more air time but less range,
  • a shallower angle gives more horizontal speed but far less time in the air.
• The trade-off is why $45°$ wins: it is the angle that balances a large $v_x$ against a long $t$.

Change the launch angle and speed below, then press Launch. Watch the range, maximum height and time of flight respond — and confirm for yourself that $30°$ and $60°$ land together:

Velocity at any point in the flight

• Horizontal component: always $v_x$.
• Vertical component at time $t$: $v_{fy} = v_y - gt$ (taking up as positive).
• Combine for the actual velocity:

$$v = \sqrt{v_x^2 + v_{fy}^2}, \qquad \tan\theta = \frac{v_{fy}}{v_x}$$

• At the highest point the velocity is not zero — the vertical component is zero, but the object is still moving horizontally at $v_x$.