Irodov 1.29: Reading a Whole Flight from One Launch – Amit & Deepali Sinha Foundation

Problem Irodov 1.29

A body is launched from ground level with initial speed $v_0$ at an angle $\alpha$ above the horizontal, under constant gravitational acceleration $g$ and negligible air drag. Find:

the total time of flight $\tau$ , until the body returns to launch height
the maximum height $h$ and the horizontal range $l$ , and the angle $\alpha$ at which $h=l$
the trajectory equation $y(x)$ , with $x$ horizontal and $y$ vertical displacement from the launch point
the radius of curvature of the trajectory at the launch point and at the highest point

The trajectory from launch to landing. v_0 leaves at angle α; the dashed droppers mark the peak height h and the range l. — The trajectory from launch to landing. $\vect v_0$ leaves at angle $\alpha$ ; the dashed droppers mark the peak height $h$ and the range $l$ .

Given

$v_0$
$\alpha$
$g$

Find

(a)

\tau

, (b)

h

l

, and the

\alpha

where

h=l

, (c)

y(x)

, (d)

R_1

at launch,

R_2

at the peak

Solution

Step 1Set up the components

With $x$ horizontal and $y$ vertical from the launch point, the horizontal velocity stays constant while gravity acts only along $y$ ; both coordinates follow from one integration of constant acceleration.

x(t)=v_0\cos\alpha\,t, \qquad y(t)=v_0\sin\alpha\,t-\frac{g}{2}t^2 \tag{1}

Step 2Flight time from the landing condition

The flight ends when $y$ returns to zero; factoring out $t$ isolates the nonzero root.

\begin{align*}0 &= v_0\sin\alpha\,\tau-\frac{g}{2}\tau^2 \\ 0 &= \tau\left(v_0\sin\alpha-\frac{g}{2}\tau\right)\end{align*}

$\tau=0$ is the launch instant; the landing is the other root.

\tau = \dfrac{2v_0\sin\alpha}{g}

(2)

Step 3Peak height and range

The peak occurs at $t=\tau/2$ , where $y(t)$ is largest; the range is $x$ evaluated at the full flight time (2).

\begin{align*}h &= y\!\left(\frac{\tau}{2}\right) \\ &= v_0\sin\alpha\cdot\frac{v_0\sin\alpha}{g}-\frac{g}{2}\left(\frac{v_0\sin\alpha}{g}\right)^2 \\ &= \frac{v_0^2\sin^2\alpha}{g}-\frac{v_0^2\sin^2\alpha}{2g} \\ l &= x(\tau) \\ &= v_0\cos\alpha\cdot\frac{2v_0\sin\alpha}{g} \\ &= \frac{2v_0^2\sin\alpha\cos\alpha}{g}\end{align*}

Collecting the two terms of $h$ over the common denominator $2g$ , and folding $l$ ’s product of a sine and a cosine into the double angle $\sin2\alpha=2\sin\alpha\cos\alpha$ ,

h=\dfrac{v_0^2\sin^2\alpha}{2g},\qquad l=\dfrac{v_0^2\sin2\alpha}{g}

(3)

Setting $h=l$ gives the angle at which the peak height equals the range.

\begin{align*}\frac{v_0^2\sin^2\alpha}{2g} &= \frac{v_0^2\sin2\alpha}{g} \\ \sin^2\alpha &= 2\sin2\alpha \\ &= 4\sin\alpha\cos\alpha \\ \frac{\sin^2\alpha}{\sin\alpha\cos\alpha} &= \frac{4\sin\alpha\cos\alpha}{\sin\alpha\cos\alpha} \\ \tan\alpha &= 4\end{align*}

\alpha = \arctan4 \approx 76^\circ

(4)

Step 4Eliminate

t

for the trajectory

Solving the horizontal law of (1) for $t$ and substituting it into the vertical law gives the path as a function of $x$ alone.

\begin{align*}t&=\frac{x}{v_0\cos\alpha} \\ y &= v_0\sin\alpha\cdot\frac{x}{v_0\cos\alpha}-\frac{g}{2}\left(\frac{x}{v_0\cos\alpha}\right)^2\end{align*}

y = x\tan\alpha-\dfrac{g}{2v_0^2\cos^2\alpha}x^2

(5)

Step 5Curvature radius from the normal component of gravity

The curvature radius of any path satisfies the general relation below, where $w_n$ is the component of acceleration perpendicular to the velocity; only gravity acts here, so $w_n$ is $g$ resolved against the local flight direction.

R = \frac{v^2}{w_n} \tag{6}

At launch the velocity makes angle $\alpha$ with the horizontal, so the component of $\vect g$ normal to $\vect v_0$ is $g\cos\alpha$ while $v=v_0$ ; at the peak the velocity is purely horizontal, so all of $\vect g$ is normal to it and $v=v_0\cos\alpha$ .

\begin{align*}w_n(0) &= g\cos\alpha, & v(0) &= v_0 \\ w_n(\tau/2) &= g, & v(\tau/2) &= v_0\cos\alpha\end{align*}

R_1=\dfrac{v_0^2}{g\cos\alpha},\qquad R_2=\dfrac{v_0^2\cos^2\alpha}{g}

(7)

Check

At $\alpha=90^\circ$ (straight up), $R_1\to\infty$ : gravity is then purely tangential at launch, decelerating the body along a straight line with no curvature, exactly as (6) predicts. Also $R_2/R_1=\cos^3\alpha\le1$ always, consistent with the peak being the tightest turn on this trajectory.

Remark

$R_1$ and $R_2$ never need the flight time, the height or the range: the curvature radius depends only on the local velocity and the local normal acceleration, which is why the same $R=v^2/w_n$ reasoning carries over unchanged to circular motion, banked curves, or any other path once the normal acceleration is identified.