Projectile Motion Equations and Formulas

Projectile motion is the curved path followed by any object that is launched into the air and then moves under the influence of gravity alone. A thrown ball, a kicked soccer ball, and a launched water balloon are all projectiles once they leave contact with whatever launched them. From that moment on, the only force acting on the object is gravity, so the object accelerates downward at a constant rate \( g \), usually taken as \( 9.8 \) meters per second squared near Earth's surface, while nothing pushes or pulls it sideways.

Because gravity only pulls straight down, the path traced out by a projectile is always a parabola (ignoring air resistance). That shape is the key to why projectile motion problems are solved by treating the horizontal and vertical directions completely separately.

The single most important idea in this topic is that projectile motion is really two simple motions happening at the same time, independently of each other:

• Horizontal direction: no force acts, so horizontal velocity \( v_x \) stays constant the entire flight.
• Vertical direction: gravity acts, so vertical velocity \( v_y \) changes at a constant rate \( g \), just like an object in free fall.

If a projectile is launched with initial speed \( v_0 \) at an angle \( \theta \) above the horizontal, the initial components are found with basic trigonometry:

\( v_x = v_0\cos\theta \)

\( v_{y0} = v_0\sin\theta \)

Once these components are known, each direction can be handled with the ordinary kinematics equations for constant acceleration, using \( a = 0 \) horizontally and \( a = -g \) vertically. This is the same component-splitting approach used when comparing motion measured from different frames, such as in the relative velocity formula.

With the components separated, the position of the projectile at any time \( t \) is described by:

\( x = v_0\cos\theta \cdot t \)

\( y = v_0\sin\theta \cdot t - \dfrac{1}{2}gt^2 \)

The vertical velocity at any time is \( v_y = v_0\sin\theta - gt \), while the horizontal velocity stays \( v_x = v_0\cos\theta \) throughout the flight. These equations for projectile motion are just the standard kinematics equations applied separately to each direction, so no new physics is needed, only careful bookkeeping of which equation applies to which direction.

Three quantities come up in almost every projectile motion problem. For a projectile launched and landing at the same height:

Time of flight: \( t = \dfrac{2v_0\sin\theta}{g} \)

Maximum height: \( h = \dfrac{v_0^2\sin^2\theta}{2g} \)

Range (horizontal distance): \( R = \dfrac{v_0^2\sin(2\theta)}{g} \)

The range formula shows that, for a fixed launch speed, distance is maximized when \( \sin(2\theta) = 1 \), which happens at \( \theta = 45^\circ \). Angles that add up to \( 90^\circ \), such as \( 30^\circ \) and \( 60^\circ \), produce the same range as each other but different flight times and maximum heights.

A ball is launched at \( 20 \) meters per second at an angle of \( 45^\circ \) above the horizontal from ground level. Find the time of flight, maximum height, and range.

Components: \( v_x = 20\cos 45^\circ \approx 14.14 \) and \( v_{y0} = 20\sin 45^\circ \approx 14.14 \), both in meters per second.

Time of flight: \( t = \dfrac{2(20)\sin 45^\circ}{9.8} \approx 2.89 \) seconds.

Maximum height: \( h = \dfrac{20^2\sin^2 45^\circ}{2(9.8)} \approx 10.2 \) meters, reached at half the total flight time.

Range: \( R = \dfrac{20^2\sin(90^\circ)}{9.8} \approx 40.8 \) meters.

Plotting height against horizontal distance for this launch produces the characteristic symmetric parabola of projectile motion, rising to the peak height near the midpoint of the range and returning to zero at landing.

Notice that the curve is symmetric: the rising half and falling half take equal time and cover equal horizontal distance, a direct result of the constant horizontal velocity and the constant downward acceleration described earlier.

Two errors show up repeatedly in projectile motion problems. The first is mixing horizontal and vertical quantities inside the same equation, such as using the launch speed \( v_0 \) directly in a vertical equation instead of its component \( v_0\sin\theta \). The second is forgetting that the range and maximum height formulas above only apply when the launch height equals the landing height; if a projectile is launched from a cliff or lands on a raised platform, the position equations \( x = v_0\cos\theta \cdot t \) and \( y = v_0\sin\theta \cdot t - \dfrac{1}{2}gt^2 \) must be solved directly instead.

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