Projectile motion: horizontal launch

What a projectile is

• A projectile is any object moving freely under gravity alone after being launched.
• Air resistance is ignored, so the only force acting is weight, straight down.
• Because the only force is vertical, the motion splits into two parts that do not affect each other.

The independence of horizontal and vertical motion

This is the key idea of the whole topic:

• Horizontally there is no force, so there is no acceleration — the horizontal velocity stays constant for the whole flight.
• Vertically the object accelerates downward at $9.8$ m s⁻², exactly like an object in free fall.
• The only quantity the two directions share is time — both motions run for the same number of seconds.

• A ball rolled off a table and a ball dropped from the table edge at the same instant hit the floor together. The horizontal velocity of the first has no effect on how fast it falls.

Setting the problem out in two columns

Always split the given information before calculating:

	Horizontal	Vertical
Acceleration	$a = 0$	$a = 9.8$ m s−2 down
Initial velocity	$v_x$ = launch speed	$v_{iy} = 0$ (horizontal launch)
Equation to use	$d_x = v_x t$	the four equations of motion

• For a horizontal launch, the initial vertical velocity is zero — the object is not thrown up or down at all.
• The launch speed is entirely horizontal.

The standard order of solving

Nearly every horizontal-launch question is solved in the same three steps:

• Find the time of flight from the vertical motion.
  • Use the drop height with $d = \tfrac{1}{2}gt^2$.
  • The horizontal speed is irrelevant here.
• Carry the time across to the horizontal motion.
  • Find the range with $d_x = v_x t$.
• Find the final velocity components if asked.
  • Horizontal: unchanged, still $v_x$.
  • Vertical: $v_{fy} = gt$.

Combining the components at landing

• The actual velocity at any instant is the vector sum of the two components:

$$v = \sqrt{v_x^2 + v_y^2}$$

• The angle below the horizontal comes from:

$$\tan\theta = \frac{v_y}{v_x}$$

• $v_x$ — horizontal component (m s⁻¹), constant

• $v_y$ — vertical component (m s⁻¹), growing throughout the flight

• $\theta$ — angle of the velocity below the horizontal

• Because $v_x$ stays fixed while $v_y$ grows, the path gets steeper and steeper — which is why the trajectory is a curve, not a straight diagonal.