$D=v_0\,t\sqrt{2(1-\sin\theta)}=25\times1.70\times\sqrt{2(1-\sin60^\circ)}\approx22\ \text{m}$
(ii) Problem Statement
"Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of $\theta=60^\circ$ to the horizontal. The initial velocity of each body is equal to $v_0=25$ m/s. Neglecting the air drag, find the distance between the bodies $t=1.70$ s later."
Reference Answer Verification (from the supplied solution image):
The reference calls the upward body 1 and the other body 2. In the frame of body 2 there is no relative acceleration, so $\vec r_{12}=\vec v_{0(12)}\,t$. Both speeds equal $v_0$, and the angle between the two launch velocities is $\pi/2-\theta$. So $v_{0(12)}=\sqrt{v_0^2+v_0^2-2v_0^2\cos(\pi/2-\theta)}$ and $|\vec r_{12}|=v_0\sqrt{2(1-\sin\theta)}\,t=22$ m. Our work matches.
(iii) The Big Idea in Plain Words
Two stones leave the same spot at the same moment. Gravity pulls both of them down. It pulls them down by the same amount. So gravity does not change the gap between them.
Symbols: $v_0$ is the launch speed. $\theta$ is the angle of body 2 above the horizontal. $g$ is the pull of gravity, $9.8$ m/s². $t$ is the time since launch. $\vec u$ is the velocity of one body as seen from the other. $D$ is the distance between the bodies.
Imagine gravity switched off. Each body would fly in a straight line at steady speed. The gap would grow at a steady rate $u$. Switching gravity back on drops both bodies by the same $\tfrac12gt^2$. The gap stays the same. So $D=u\,t$.
(iv) The Solution Viewed in Each Reference Frame
Numbers used: $v_0=25$ m/s, $\theta=60^\circ$, $t=1.70$ s, $g=9.8$ m/s². Then $\sin60^\circ=0.8660$ and $\sqrt{2(1-\sin60^\circ)}=0.5176$.
1. Ground or Lab Frame (Observer at the Launch Point)
Origin fixed at O, the launch pointWhy: pick $x$ along the ground and $y$ straight up. Then write where each body is at time $t$.
Body 1: $x_1=0$, $y_1=v_0t-\tfrac12gt^2$. Body 2: $x_2=v_0\cos\theta\,t$, $y_2=v_0\sin\theta\,t-\tfrac12gt^2$.
Why: put in the numbers. First find the drop $\tfrac12gt^2=\tfrac12(9.8)(1.70)^2=14.16$ m.
$y_1=42.50-14.16=28.34$ m. $x_2=25(0.5)(1.70)=21.25$ m. $y_2=36.81-14.16=22.65$ m.
Why: the gap is the straight line from body 1 to body 2. Find its sideways and up-down parts first.
$\Delta x=21.25-0=21.25$ m. $\Delta y=22.65-28.34=-5.69$ m. The drop $\tfrac12gt^2$ cancelled in $\Delta y$.
Why: Pythagoras turns the two parts into one distance.
$$D=\sqrt{\Delta x^2+\Delta y^2}=\sqrt{21.25^2+5.69^2}=\sqrt{483.98}=22.0\ \text{m}$$
In symbols: $D=v_0t\sqrt{\cos^2\theta+(1-\sin\theta)^2}=v_0t\sqrt{2(1-\sin\theta)}$.
Why: a velocity check shows the gap grows at a steady rate.
At $1.70$ s: $\vec v_1=(0,\ 8.34)$ m/s and $\vec v_2=(12.50,\ 4.99)$ m/s. So $\vec v_2-\vec v_1=(12.50,\ -3.35)$ m/s. This is the same at every time. Its size is $12.94$ m/s.
2. Falling-Body Frame (Observer Riding Body 2, the Book's Frame)
Origin moves with body 2 and speeds up down at $g$Why: this frame falls with acceleration $g$. A rider feels a made-up force called a pseudo force. It is $mg$ upward on every body. It exactly cancels the weight.
So no net force acts on any body here. Each body moves in a straight line at steady speed.
Why: body 2 sits at the origin, so we only need the motion of body 1.
Body 1 has velocity $\vec v_1-\vec v_2=(-v_0\cos\theta,\ v_0(1-\sin\theta))=(-12.50,\ +3.35)$ m/s. This never changes.
Why: the speed of body 1 here comes from the triangle of the two launch velocities. They make an angle of $90^\circ-\theta=30^\circ$.
$u=\sqrt{v_0^2+v_0^2-2v_0^2\cos30^\circ}=v_0\sqrt{2-\sqrt3}=25\times0.5176=12.94$ m/s.
Why: steady speed means distance is speed times time.
$D=u\,t=12.94\times1.70=22.0$ m. In body 1's frame, body 2 moves at $\vec v_2-\vec v_1$: the same size, the opposite direction. Same $D$.
3. Centre-of-Mass (CM) Frame
Origin moves with the CM (equal masses drawn)Why: the CM is the balance point of the two bodies. The problem gives no masses. The gap does not depend on them, so we draw equal masses $m$.
The CM velocity is $\vec v_{cm}=\tfrac12(\vec v_1+\vec v_2)$. The CM also falls with acceleration $g$.
Why: this frame accelerates at $g$ downward. So each body feels a pseudo force $mg$ upward. It cancels the weight, and the motion is straight lines.
Body 1 moves at $\vec v_1-\vec v_{cm}=-\tfrac12\vec u$. Body 2 moves at $+\tfrac12\vec u$. Each speed is $\tfrac12u=6.47$ m/s.
Why: the total momentum in the CM frame must be zero. Check it.
$m(-\tfrac12\vec u)+m(+\tfrac12\vec u)=0$. It is zero at every time, for any masses.
Why: the bodies move apart in opposite directions from the CM. Add their two distances.
$D=\tfrac12ut+\tfrac12ut=ut=22.0$ m. Each body is $11.0$ m from the CM.
4. Medium (Air) Frame
This frame does not apply. Air drag is neglected, so there is no air to ride with.
What Changes and What Stays the Same Across Frames
- What changes: the paths and the velocities. On the ground each body follows a curve called a parabola. In the falling frames the paths are straight lines. A pseudo force appears and cancels gravity.
- What stays the same: the relative velocity $\vec u$, its size $12.94$ m/s, the time, and the gap $D=ut=22.0$ m.
(v) The Solution by Each Method
Method 1: Relative-Motion Method
Fastest- Why: both bodies have the same acceleration $g$, so the relative acceleration is zero. The relative velocity is constant, and $\vec r_{12}=\vec u\,t$.
- Why: find the angle between the two launch velocities. Body 1 points up. Body 2 is $\theta$ above the ground.
The angle is $90^\circ-60^\circ=30^\circ$. - Why: the law of cosines gives the size of $\vec u$ from two equal sides $v_0$ and the angle between them.
$u=\sqrt{2v_0^2(1-\cos30^\circ)}=v_0\sqrt{2(1-\sin60^\circ)}=12.94$ m/s. - Why: distance is speed times time.
$D=12.94\times1.70\approx22$ m.
Method 2: Force Method
- Why: the free-body diagram shows what pushes each body. Only the weight acts (red arrows). Air drag is neglected.
- Why: Newton's second law gives each acceleration.
$m\vec a=m\vec g\Rightarrow\vec a_1=\vec a_2=\vec g$. - Why: subtract the two to see how the gap behaves.
$\vec a_2-\vec a_1=0$. So $\vec v_2-\vec v_1=\vec v_{02}-\vec v_{01}=\vec u$ is constant. - Why: with a constant $\vec u$, the gap vector is $\vec u\,t$.
$\vec u=(12.50,\ -3.35)$ m/s, so $D=\sqrt{12.50^2+3.35^2}\times1.70=12.94\times1.70\approx22$ m.
Method 3: Graph Method (Gap Against Time)
- Why: the gap grows at the steady rate $u$. So the graph of $D$ against $t$ is a straight line through the origin. The graph at the bottom of the simulation draws it.
- Why: a table of values shows the pattern.
$t=0.5$ s: $6.47$ m. $\ 1.0$ s: $12.94$ m. $\ 1.5$ s: $19.41$ m. $\ 1.7$ s: $22.00$ m. $\ 2.0$ s: $25.88$ m. - Why: the slope of the line is the relative speed. Its steady value also means the area under the $u$–$t$ graph is a rectangle.
Slope $=12.94$ m/s. Area up to $1.70$ s $=12.94\times1.70\approx22$ m. Read $D$ at $t=1.70$ s on the line: $22$ m.
Method 4: Calculus Method
- Why: start from the acceleration, which is the same for both bodies.
$\dfrac{d^2\vec r}{dt^2}=\vec g$ for body 1 and body 2. - Why: integrate once for velocity and again for position, using the start values.
$\vec r_i=\vec v_{0i}\,t+\tfrac12\vec g\,t^2$, since both start at the origin. - Why: subtract. The gravity term is the same in both, so it cancels.
$\vec r_{12}=\vec r_2-\vec r_1=(\vec v_{02}-\vec v_{01})\,t=\vec u\,t$. - Why: take the size and put in the numbers.
$D=t\sqrt{(v_0\cos\theta)^2+(v_0\sin\theta-v_0)^2}=v_0t\sqrt{2(1-\sin\theta)}\approx22$ m. Also $\dfrac{dD}{dt}=u$ is constant, so $D$ has no maximum or minimum.
Method 5: Energy Method
Not suited. Energy gives the speed of each body, but not the direction. The gap needs directions, so energy alone cannot find $D$.
Fastest: Method 1 (relative motion), a two-line answer. Most reliable in an exam: Method 4, or the ground-frame components, because you can check every number. All give $D\approx22$ m.
(vi) Parameter Table
| Symbol | Meaning | Unit | Live value |
|---|---|---|---|
| $v_0$ | launch speed of each body | m/s | |
| $\theta$ | angle of body 2 above the horizontal | deg | |
| $g$ | pull of gravity (cancels in $D$) | m/s² | |
| $t$ | time since launch | s | |
| $\alpha$ | angle between the two launch velocities, $90^\circ-\theta$ | deg | |
| $u$ | relative speed, $v_0\sqrt{2(1-\sin\theta)}$ | m/s | |
| $t_{land,2}$ | when body 2 would land, $2v_0\sin\theta/g$ | s | |
| $D$ | distance between the bodies, $ut$ (answer) | m |
(vii) Common Exam Traps
- Using $\cos\theta$ in the law of cosines. The angle between the velocities is $90^\circ-\theta=30^\circ$, not $60^\circ$. That is why the result has $\sin\theta$.
- Using the wrong angle gives $v_0\sqrt{2(1-\cos60^\circ)}\,t=42.5$ m. That is almost double the right answer.
- Forgetting that "straight up" means $90^\circ$ to the horizontal.
- Adding the two speeds, $v_0+v_0$. Velocities are vectors, so subtract them as vectors.
- Thinking gravity changes the gap. It does not. The drop $\tfrac12gt^2$ is the same for both bodies and cancels.
- Giving only the sideways gap ($21.25$ m) or only the up-down gap ($5.69$ m). The question asks for the straight-line distance.
- Not checking that both bodies are still in the air. Body 2 lands at $4.42$ s, so $1.70$ s is safe. After landing, the formula no longer holds.
- Rounding too early. Using $0.52$ instead of $0.5176$ gives $22.1$ m. Keep four digits until the end.
- Mixing units. Speed in m/s times time in s gives metres.
Verified: the live numbers, all four frames and all methods give $D\approx22$ m for $v_0=25$ m/s, $\theta=60^\circ$, $t=1.70$ s, with speeds in m/s, times in s and lengths in m. The CM block assumes equal masses for the drawing only.