In a school ground two friends throw a ball each straight up at the same instant, one harder than the other. Both balls rise, slow down, stop and fall back. Each ball changes its velocity all the time. Yet a photograph taken every half second shows the gap between the balls growing by the same amount each time.
Gravity drops out
Both balls are in free fall: after they leave the hands, the only acceleration of each is \(g\), downward. Take upward as positive. Then \(a_1 = a_2 = -g\), and the relative acceleration is \(a_{12} = -g - (-g) = 0\). With no relative acceleration, the relative velocity keeps its starting value, \(u_1 - u_2\).
\[ y_1 = u_1 t - \tfrac{1}{2} g t^2, \qquad y_2 = u_2 t - \tfrac{1}{2} g t^2 \]
Why. the second equation of motion for each ball, thrown upward from the same point at the same instant, with upward positive.
\[ y_1 - y_2 = u_1 t - u_2 t - \tfrac{1}{2} g t^2 + \tfrac{1}{2} g t^2 \]
Why. subtract the two positions to get the position of ball 1 relative to ball 2.
\[ y_1 - y_2 = (u_1 - u_2)\,t \]
Read it. the two gravity terms are equal and cancel. The separation grows in proportion to the time.
Watch the two balls in Fig 5.1 and the graph beside them. Ball 2 turns at \(2\,\text{s}\). Ball 1 turns at \(3\,\text{s}\). After \(3\,\text{s}\) both balls fall, both speed up, and still the separation grows by exactly \(10\,\text{m}\) every second. Each blue arrow changes by \(10\,\text{m/s}\) every second, but the two change together, so their difference stays \(10\,\text{m/s}\).
When two bodies move under gravity alone, near the Earth and without air resistance, the motion of one as seen from the other is uniform. It is motion in a straight line at the constant relative velocity with which they started.
Meeting in the air
The same idea solves the meeting problem at once. A stone is dropped from the top of a tower. At the same instant another stone is thrown up from the foot. In the frame of the dropped stone there is no gravity, and the second stone comes straight up at its launch speed. So the stones meet after the height of the tower divided by that speed.
A stone is dropped from the top of a tower \(80\,\text{m}\) high. At the same instant a second stone is thrown straight up from the foot of the tower at \(40\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). When and where do the stones meet?
Upward positive. Stone 1 (dropped): \(u_1 = 0\). Stone 2 (thrown): \(u_2 = +40\,\text{m/s}\). Both accelerate at \(-10\,\text{m/s}^2\). The initial separation is \(80\,\text{m}\).
The velocity of stone 2 relative to stone 1 is \(u_2 - u_1 = +40\,\text{m/s}\), constant. The gap of \(80\,\text{m}\) closes at \(40\,\text{m/s}\).
\(t = 80 / 40 = 2\,\text{s}\). The dropped stone has fallen \(\tfrac{1}{2}(10)(2^2) = 20\,\text{m}\), so the meeting point is \(60\,\text{m}\) above the ground.
The stones meet after \(2\,\text{s}\), \(60\,\text{m}\) above the ground and \(20\,\text{m}\) below the top. The value of \(g\) never entered the meeting time.
A ball is dropped from a height of \(45\,\text{m}\). At the same instant a second ball is thrown straight up from the ground below it at \(20\,\text{m/s}\). After how many seconds do the balls meet?
Two balls are thrown up at the same instant from the same point, one at \(15\,\text{m/s}\) and one at \(10\,\text{m/s}\). While both are in the air, how does the faster ball move as seen from the slower ball?
In 1907 Einstein was a clerk in the patent office at Bern. One day, he later said, a thought struck him: a person falling freely from a roof would not feel their own weight. A stone let go during the fall would stay beside the person, at rest or moving steadily. Seen from the falling person, gravity has disappeared, exactly as it disappears between the two balls of this lesson. Einstein called it the happiest thought of his life. Eight years of work turned it into the general theory of relativity.
The gap between two lines
Draw the x–t graphs of two bodies on the same axes. At any time \(t\), read both lines. The vertical gap from B's line up to A's line is \(x_A - x_B\), the relative position \(x_{AB}\). The slope of each line is a velocity, so the difference of the slopes is \(v_{AB}\).
Press play in Fig 5.2 with "A catches B". A starts at \(0\) with \(10\,\text{m/s}\), and B starts at \(30\,\text{m}\) with \(5\,\text{m/s}\). The gap starts at \(-30\,\text{m}\) and shrinks by \(5\,\text{m}\) every second. At \(6\,\text{s}\) the lines cross, \(x_{AB} = 0\), and A passes B. Try the other chips. Lines that come together mean the bodies approach. Parallel lines would mean a constant gap and \(v_{AB} = 0\).
The same reading works for any pair of graphs. On two v–t graphs, the vertical gap is \(v_{AB}\), and the difference of the slopes is \(a_{AB}\). For the two balls of Fig 5.1, the v–t graphs are two parallel lines of slope \(-g\). The gap between them is the constant \(u_1 - u_2\), and that is the whole of the physics of this lesson in one picture.
Two bodies move along the x axis with \(x_A = 4t\) and \(x_B = 24 - 2t\), where \(x\) is in metres and \(t\) in seconds. At what time do their x–t lines cross?
The x–t graphs of two cars on a straight road are two parallel straight lines, one above the other. What does this tell you?
- Keeping \(g\) in the relative motion of two freely falling bodies.Both have acceleration \(g\), so the relative acceleration is zero. Use the relative velocity alone.
- Reading the gap between two x–t lines horizontally.The relative position at a given time is the vertical gap, at the same \(t\).
- Assuming that a dropped stone and a stone thrown up from below meet halfway.They meet where the gap has closed at the relative speed. Find the time first, \(t = h/u\), then the height.