Projectile Motion Analysis

Start by Splitting the Initial Velocity

Projectile analysis almost always starts with one simple question: how much of the velocity is horizontal, and how much is vertical? If the initial speed is $v_0$ and the launch angle is $\theta$ , the initial components are:

v_{0x}=v_0\cos\theta

After that, the horizontal and vertical directions are calculated with different rules. The horizontal direction uses uniform motion because there is no horizontal acceleration. The vertical direction uses uniformly accelerated motion because gravity keeps acting downward.

The key to projectile analysis is using the same time for two different directions.

Cannonball Component Analysis

A cannonball crosses the water while the ghost balls mark equal time steps from the same formulas used in the calculation.

Horizontal component: $v_{0x}=20\text{ m/s}$
Vertical component: $v_{0y}=34.6\text{ m/s}$
Peak time: $t=3.5\text{ s}$
Flight time: $T=6.9\text{ s}$
Range: $R=138.6\text{ m}$
Velocity at two seconds: $\vec{v}=\langle 20, 14.6\rangle\text{ m/s}$

The arc and ghost balls show positions at equal time intervals, not separate horizontal and vertical paths. The values below the scene come from the same split: $v_{0x}$ stays constant, while $v_y=v_{0y}-gt$ changes because gravity acts downward.

The number flow is easier to read as one connected formula sequence:

\begin{aligned} v_{0x}&=v_0\cos\theta\\ v_{0y}&=v_0\sin\theta\\ t_{\text{peak}}&=\frac{v_{0y}}{g}\\ T&=2t_{\text{peak}}\\ R&=v_{0x}T\\ \vec{v}(2)&=\langle v_{0x},v_{0y}-g(2)\rangle,\quad g=10\text{ m/s}^2 \end{aligned}

So the values below the scene are not manually chosen numbers. They are calculated from the selected launch, then shown as components, time, range, and instantaneous velocity.

The Peak Happens When Vertical Velocity Is Zero

At the peak, the object is still moving horizontally, but its up-and-down motion pauses for an instant. So the peak condition is $v_y=0$ .

v_y=v_{0y}-gt

When $v_y=0$ , the time to the peak is:

t_{\text{peak}}=\frac{v_{0y}}{g}

For an initial speed of $40\text{ m/s}$ at an angle of $60^\circ$ , the initial vertical component is $20\sqrt{3}\text{ m/s}$ . Before the peak it is positive, at the peak it becomes zero, and after the peak it is negative because the object is moving downward.

Maximum Range Uses Total Time

If the object launches from the ground and returns to the ground at the same height, the total time is twice the time to the peak.

t_{\text{total}}=2t_{\text{peak}}

This formula shows why range is not determined by angle alone. Range also depends on horizontal velocity and how long the object stays in the air.

Instantaneous Velocity Comes from Two Components

To find velocity at a specific time, calculate $v_x$ and $v_y$ at that time first. The horizontal component stays constant, while the vertical component changes.

v_x=v_{0x}

The direction of velocity can also be read from the ratio of the vertical and horizontal components:

\tan\alpha=\frac{v_y}{v_x}

If $v_y$ is positive, the velocity points upward. If $v_y$ is negative, the velocity points downward.

A Sixty Degree Launch Sets Two Components

A ball is launched with an initial speed of $40\text{ m/s}$ and an angle of $60^\circ$ . Use $g=10\text{ m/s}^2$ .

This example uses an angle with clear trigonometric components, so the full analysis stays visible: split the initial velocity, get time from vertical motion, then use the same time to read range and instantaneous velocity.

The initial components are:

v_{0x}=40\cos60^\circ=20\text{ m/s}

The time to the peak is:

t_{\text{peak}}=\frac{20\sqrt{3}}{10}=2\sqrt{3}\text{ s}

The total time and range are:

t_{\text{total}}=4\sqrt{3}\text{ s}

Suppose we want the velocity at $t=2\text{ s}$ . The vertical component is:

v_y=20\sqrt{3}-10(2)=20\sqrt{3}-20\text{ m/s}

Meanwhile, $v_x=20\text{ m/s}$ stays constant. The speed is:

\begin{aligned} v&=\sqrt{20^2+(20\sqrt{3}-20)^2}\\ &=\sqrt{2000-800\sqrt{3}}\\ &\approx 24.8\text{ m/s} \end{aligned}

The direction of the velocity from the horizontal is:

\alpha=\tan^{-1}\left(\frac{20\sqrt{3}-20}{20}\right)\approx 36.2^\circ

So at $t=2\text{ s}$ , the ball is still moving upward at an angle because $v_y$ is still positive.