Projectile Motion: How to Split Horizontal and Vertical to Solve Any Throw

Projectile motion problems look complicated because there is a lot happening at once. The trick that makes every single one tractable is this: horizontal and vertical motion are completely independent. Gravity only affects the vertical direction. The horizontal component of velocity never changes. Treat them as two separate problems that share only the time variable.

The two sets of equations

• Horizontal: $x=v_0\cos (\theta )$· t. No acceleration. Constant velocity.
• Vertical: $y=v_0\sin (\theta )$· $t-\frac{1}{2}g{t}^2$. $Acceleration=-g=-9.81\frac{m}{s^2}$ downward.
• Vertical velocity: vᵧ $=v_0\sin (\theta )-g$·t.

At the peak: vertical velocity is zero

The projectile reaches its maximum height when vᵧ $=0$. Set $v_0\sin (\theta )-g$·$t=0$ and $solve:t_{peak}=v_0\frac{\sin (\theta )}{g}$. The total time of flight is 2 · $t_{peak}(assumingflatground)$. The horizontal range is $x=v_0\cos (\theta )$· 2t_peak.

Worked example

A ball is launched at 20 m/s at 30° above horizontal. Find time of flight, range, and max height. $t_{peak}$ $=20$· sin(30°$)/9.81=20$$\cdot \frac{0.5}{9.81}\approx 1.02s$. Total $time=2.04s$. $Range=20$· cos(30°) · $2.04=20$$\cdot 0.866\cdot 2.04\approx 35.4m.$ $Maxheight=v_0\sin (30$°) · $t_{peak}-\frac{1}{2}$· 9.81 · $t_{peak}^2=10$$\cdot 1.02-4.905\cdot 1.04\approx 5.1m.$

The three common mistakes

• Using the full launch speed instead of its component. Always decompose v₀ into v₀ cos(θ) horizontally and v₀ sin(θ) vertically before writing any equation.
• Using $g=9.81$ with the wrong sign. Gravity acts downward. If upward is positive, the vertical acceleration is -9.81 m/$s^2$. Getting this sign wrong flips the entire solution.
• Treating the two axes as linked. Horizontal speed does not change during flight; the ball does not slow down horizontally because gravity is pulling it down.

When the ground is not flat

If the launch height differs from the landing height, the total flight time is no longer simply 2·$t_{peak}$. Set $y=y_{landin}g(e.g.,y=-h$ for a cliff problem) and solve the resulting quadratic for t. There will be two solutions; only the positive one is physical.

Maximum range is at 45° — but only on flat ground

For launch and landing at the same height, range = v₀₂ sin(2θ) / g is maximized at θ = 45°. The sin(2θ) term shows that complementary angles (e.g., 30° and 60°) give identical range. This symmetry is a useful sanity check on your answer.

Upload a projectile problem to Solvequill and the whiteboard video decomposes the vector, labels the equations on each axis, and shows each substitution step. It is particularly useful for seeing which component to reach for first.