Irodov 1.27: Reading Launch Speed Off a Parabola's Shape

Problem Irodov 1.27

A particle moves in the $xy$ plane under a constant acceleration $\vect w$ of magnitude $w$ , directed along the negative $y$ axis. Its trajectory is the parabola $y=ax-bx^2$ , with $a$ and $b$ known positive constants. Find the particle’s velocity (magnitude) at the instant it passes through the origin.

The parabolic path through the origin. The launch velocity v_0 at O is tangent to the path there; the constant acceleration w points along - j throughout the flight. — The parabolic path through the origin. The launch velocity $\vect v_0$ at $O$ is tangent to the path there; the constant acceleration $\vect w$ points along $-\vect j$ throughout the flight.

Given

$y=ax-bx^2$
$\vect w\parallel-\vect j$ , $|\vect w|=w$

Find

v_0

at the origin

Solution

Step 1Write the motion in components

With $\vect w=-w\vect j$ constant and no acceleration along $x$ , the two coordinates evolve independently from the launch instant $t=0$ at the origin, where the velocity components are $v_{0x}$ and $v_{0y}$ .

x(t) = v_{0x}t, \qquad y(t) = v_{0y}t-\frac{w}{2}t^2

Step 2Eliminate

t

and match the given trajectory

Solving the $x$ equation for $t$ and substituting into $y(t)$ expresses the path as a function of $x$ alone, which must be identical to the stated parabola for every $x$ .

\begin{align*}t &= \frac{x}{v_{0x}} \\ y &= v_{0y}\frac{x}{v_{0x}}-\frac{w}{2}\left(\frac{x}{v_{0x}}\right)^2 \\ &= \frac{v_{0y}}{v_{0x}}\,x-\frac{w}{2v_{0x}^2}\,x^2\end{align*}

Matching coefficients with $y=ax-bx^2$ term by term gives two equations for the launch components, and they are the crux of the problem: the parabola’s two constants are exactly the two things the launch velocity has to supply.

a=\dfrac{v_{0y}}{v_{0x}},\qquad b=\dfrac{w}{2v_{0x}^2}

(1)

Step 3Solve for the launch components

The second of (1) isolates $v_{0x}$ directly, and the first of them then gives $v_{0y}$ in terms of it.

\begin{align*} v_{0x}^2 &= \frac{w}{2b} \\ v_{0x} &= \sqrt{\frac{w}{2b}} \\ v_{0y} &= a\,v_{0x} \\ &= a\sqrt{\frac{w}{2b}} \tag{2}\end{align*}

Step 4Combine into the launch speed

The speed at the origin is the root sum of squares of the two components (2).

\begin{align*}v_0^2 &= v_{0x}^2+v_{0y}^2 \\ &= \frac{w}{2b}+a^2\frac{w}{2b} \\ &= \frac{(1+a^2)w}{2b}\end{align*}

v_0 = \sqrt{\dfrac{(1+a^2)w}{2b}}

(3)

Check

Dimensionally $a$ is a plain slope and $b$ has units of inverse length, so $w/b$ carries units of length $^2$ /time $^2$ , and $\sqrt{w/2b}$ is a speed, as required. As $a\to0$ the launch is purely horizontal ( $v_{0y}=0$ ) and $v_0\to\sqrt{w/2b}=v_{0x}$ , consistent with Step 3.

Remark

The parabola’s own shape already fixes the launch velocity completely: $b$ alone sets the horizontal speed through $v_{0x}=\sqrt{w/2b}$ (how sharply the path curves, for a given $w$ ), and $a=\tan\theta_0$ is just the launch angle’s slope. This is the same reading used to time a gravity-driven projectile from its trajectory’s coefficients alone, without ever separately measuring a speed or an angle.