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. In the model used here, air resistance is ignored and gravitational acceleration is constant. 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 horizontal and vertical equations use the same time variable.

Cannonball Component Analysis

A cannonball crosses the water while faded position markers show its location at equal time intervals.

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 faded balls show one motion at equal time intervals. The values below split that motion into components: $v_{0x}$ stays constant, while $v_y=v_{0y}-gt$ changes because gravity acts downward.

The scenes launch and land at the same height. Under that condition, time of flight, maximum height, and range can be calculated in one 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, \\ &g=10\text{ m/s}^2 \end{aligned}

The selected launch determines every value below the scene. The formulas calculate the 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.

Range Uses Total Flight 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 Has Horizontal and Vertical 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 must use both components and the correct quadrant:

\alpha=\operatorname{atan2}(v_y,v_x)

If $v_x>0$ , this is equivalent to $\tan\alpha=v_y/v_x$ with $-90^\circ<\alpha<90^\circ$ . A positive $v_y$ points above the horizontal, while a negative value points below it.

Launch at Sixty Degrees

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

At $60^\circ$ , the sine and cosine have exact values. We can therefore split the initial velocity exactly, find the time from the vertical motion, and use that same time to calculate the 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:

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

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=\operatorname{atan2}(20\sqrt{3}-20,20)\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.