Irodov 1.10: Gravity Cancels, Only Launch Differs – Amit & Deepali Sinha Foundation

Problem Irodov 1.10

Two small bodies are launched at the same instant from the same point, both with the same initial speed $v_0 = 25$ m/s: body 1 straight up, body 2 at angle $\theta = 60^\circ$ above the horizontal. Both move freely under gravity, air drag neglected. Find the straight-line distance between the two bodies at time $t = 1.70$ s after launch.

The two launch velocities from the common point, both of magnitude v_0: v_1 straight up, v_2 at θ=60° to the horizontal, 30° apart. — The two launch velocities from the common point, both of magnitude $v_0$ : $\vect v_1$ straight up, $\vect v_2$ at $\theta=60^\circ$ to the horizontal, $30^\circ$ apart.

Given

$v_0 = 25$ m/s
$\theta = 60^\circ$
$t = 1.70$ s

Find

the distance

l

between the bodies at time

t

Solution

Step 1The shared acceleration cancels out of the separation

Each body’s position is its launch velocity times $t$ plus the common free-fall term, so subtracting one from the other cancels that term exactly.

\begin{align*}\vect r_1(t) &= \vect v_1 t+\frac12\vect g t^2\\ \vect r_2(t) &= \vect v_2 t+\frac12\vect g t^2\end{align*}

Subtracting the second line from the first, the common $\vect g t^2/2$ term drops out identically.

\vect r_1(t)-\vect r_2(t) = (\vect v_1-\vect v_2)\,t

(1)

Step 2The magnitude of the launch-velocity difference

Vector subtraction of the two launch velocities, head to tail: v_1 from the dot, then - v_2 from its tip; the closing arrow v_1- v_2 runs from the dot back to that endpoint, the fixed direction along which grows. — Vector subtraction of the two launch velocities, head to tail: $\vect v_1$ from the dot, then $-\vect v_2$ from its tip; the closing arrow $\vect v_1-\vect v_2$ runs from the dot back to that endpoint, the fixed direction along which (1) grows.

With $x$ horizontal and $y$ vertical, $\vect v_1=(0,v_0)$ and $\vect v_2=(v_0\cos\theta,v_0\sin\theta)$ , both of magnitude $v_0$ since the two launch speeds are equal. Their difference and its squared length follow by direct subtraction.

\begin{align*}\vect v_1-\vect v_2 &= \bigl(-v_0\cos\theta,\ v_0-v_0\sin\theta\bigr)\\ |\vect v_1-\vect v_2|^2 &= v_0^2\cos^2\theta+v_0^2(1-\sin\theta)^2\\ &= v_0^2\left[\cos^2\theta+\sin^2\theta+1-2\sin\theta\right]\\ &= 2v_0^2(1-\sin\theta)\end{align*}

|\vect v_1-\vect v_2| = v_0\sqrt{2(1-\sin\theta)}

(2)

Step 3The separation distance

Taking the magnitude of (1) and substituting (2) gives the separation as a function of time, then the given numbers evaluate it.

l(t) = |\vect r_1(t)-\vect r_2(t)| = |\vect v_1-\vect v_2|\,t

With $v_0=25$ m/s, $t=1.70$ s and $\sin 60^\circ=0.866$ , the bracket evaluates to $2(1-0.866)=0.268$ . Numerically,

l = (25)(1.70)\sqrt{0.268}\text{ m}

l = v_0 t\sqrt{2(1-\sin\theta)} = 22\text{ m}

(3)

Check

Dimensionally, $[v_0][t]=(\text{m/s})(\text{s})=\text{m}$ , a length, as required. In the limit $\theta\to90^\circ$ (both bodies thrown straight up together) the bracket $2(1-\sin\theta)\to0$ and $l\to0$ , correctly reporting no separation when the two launches coincide.

Remark

Because the $\vect g t^2/2$ term cancels identically, the two bodies stay separated by a vector that never changes direction, only length: from the instant of launch until either one lands, the line joining them is always parallel to $\vect v_1-\vect v_2$ , and the separation grows strictly linearly in $t$ . Seen from either body, the other one simply recedes in a straight line at constant speed $|\vect v_1-\vect v_2|$ , exactly as if gravity were absent.