CLASSICAL-MECHANICS · Interactive Practice — Unit 4 · Video 4

Erasing Time: How a Projectile's Path Becomes a Parabola

IKey Formulas

Formula	Name	Description
$x(t) = x_0 + v_{x,0}\,t,\quad y(t) = y_0 + v_{y,0}\,t - \tfrac{1}{2} g t^2$	Equations of motion	Horizontal and vertical position vs. time
$y = -\dfrac{1}{2}\dfrac{g}{v_{x,0}^2}\,x^2 + \left(\dfrac{g x_0}{v_{x,0}^2} + \dfrac{v_{y,0}}{v_{x,0}}\right)x + C$	Orbit equation	Path $y(x)$ after eliminating $t$
$\theta = \arctan\!\left(\dfrac{dy}{dx}\right)$	Direction of motion	Velocity angle equals the arctan of the slope
$y = \tan(\theta_0)\,x - \dfrac{g}{2 v_0^2 \cos^2\theta_0}\,x^2$	Launch-origin form	Compact orbit with origin at the launch point

IIInteractive Visualizations

Visualization 1 — Erasing Time from the Path

Sampled at equal time steps the projectile crowds near the peak, yet the shape it traces is a single downward parabola.

Launch speed v₀ (m/s) 25 Launch angle θ₀ (°) 50 Time

Visualization 2 — Slope Is the Direction of Motion

At each point the tangent's slope is $\tan\theta$ — the heading of the velocity, not its speed.

Position along path 30%

💡 Two projectiles can ride this exact tangent at different speeds — the slope fixes the heading, never the magnitude $|\vec{v}|$ .

Visualization 3 — Why the Arc Opens Downward

The $x^2$ coefficient $-\tfrac{1}{2}\,g/v_{x,0}^2$ is always negative, so smaller $v_{x,0}$ only curls the downward arc tighter.

Horizontal speed v_x,0 (m/s) 18

IIIQuiz Questions

Question 1

To eliminate time and obtain the orbit equation $y(x)$ , the strategy from the video is to solve one equation of motion for $t$ and substitute it into the other. Which equation should you solve for $t$ , and what do you get?

✅ Correct! The horizontal equation is linear in $t$ , so it solves cleanly: $t = (x - x_0)/v_{x,0}$ .

❌ Not quite. Pick the equation that is linear in $t$ so the algebra stays clean — that is the horizontal equation $x(t)$ .

Show solution

Solution:

The horizontal equation $x = x_0 + v_{x,0}\,t$ is linear in $t$ , so it solves cleanly for time:

$t = \frac{x - x_0}{v_{x,0}}.$

Substituting this single expression into the vertical equation everywhere $t$ appears eliminates time and leaves a relationship purely between $y$ and $x$ . Solving the $y$ -equation for $t$ would mean inverting a quadratic in $t$ (messy, with a square root), which is exactly what we want to avoid.

Question 2

In the expanded orbit equation, the coefficient of $x^2$ is $-\dfrac{1}{2}\dfrac{g}{v_{x,0}^2}$ . What does the sign of this coefficient tell you about the shape of the path?

✅ Correct! With $g>0$ and $v_{x,0}^2>0$ , the leading minus sign forces a negative coefficient — a downward-opening parabola.

❌ Not quite. Both $g$ and $v_{x,0}^2$ are positive, so the minus sign out front fixes the coefficient as negative — always.

Show solution

Solution:

The coefficient is

$-\frac{1}{2}\,\frac{g}{v_{x,0}^2}.$

Gravity $g > 0$ and the squared horizontal speed $v_{x,0}^2 > 0$ , so the ratio is positive and the leading minus sign makes the whole coefficient negative. A negative $x^2$ coefficient means the parabola opens downward — exactly what a thrown object does as it arcs up and falls back. The launch angle changes how stretched the curve is, but it can never flip this sign.

Question 3

True or False: The slope $\dfrac{dy}{dx}$ of the orbit at a point gives you both the direction of the velocity and the speed $|\vec{v}|$ of the projectile at that point.

✅ Correct! The slope fixes only the direction. Speed needs the actual components $v_x$ and $v_y$ , not just their ratio.

❌ Not quite. The slope is a ratio $v_y/v_x$ , which keeps the direction but throws away the magnitude — so it can't give speed.

Show solution

Solution: False.

The slope gives only the direction of motion: $\theta = \arctan(dy/dx)$ fixes the angle of the velocity. But two projectiles can trace the exact same parabola while one races along it and the other crawls — the tangent line fixes the heading, not the magnitude. To recover the speed you need the velocity components themselves, $v_x = dx/dt$ and $v_y = dy/dt$ , and then $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$ . Their ratio (the slope) discards the size information.

Shape gives you heading; it does not give you speed.

Question 4

Place the origin at the launch point so $x_0 = 0$ and $y_0 = 0$ , and write the orbit in terms of the launch speed $v_0$ and angle $\theta_0$ . What is the coefficient of the linear $x$ term in $\,y = \tan(\theta_0)\,x - \dfrac{g}{2 v_0^2 \cos^2\theta_0}\,x^2$ ?

✅ Correct! The linear coefficient is $v_{y,0}/v_{x,0} = \tan(\theta_0)$ — the launch slope.

❌ Not quite. The linear term comes from $v_{y,0}/v_{x,0}$ . Using $v_{y,0}=v_0\sin\theta_0$ and $v_{x,0}=v_0\cos\theta_0$ this ratio simplifies to $\tan(\theta_0)$ .

Show solution

Solution:

With the origin at the launch point the constant terms and the $g x_0/v_{x,0}^2$ piece all drop out, leaving

$y = -\frac{1}{2}\frac{g}{v_{x,0}^2}\,x^2 + \frac{v_{y,0}}{v_{x,0}}\,x.$

The linear coefficient is $v_{y,0}/v_{x,0}$ . Substituting $v_{x,0} = v_0\cos\theta_0$ and $v_{y,0} = v_0\sin\theta_0$ ,

$\frac{v_{y,0}}{v_{x,0}} = \frac{v_0\sin\theta_0}{v_0\cos\theta_0} = \tan\theta_0.$

So the line through the launch point has slope $\tan(\theta_0)$ — the launch slope. The other option, $g/(2 v_0^2 \cos^2\theta_0)$ , is the coefficient of the $x^2$ term, not the $x$ term.

Solved: 0 / 4