A batter lofts the ball towards the boundary. The fielder watches it for about a second and then runs, not to where the ball is, but to where it will come down. The fielder has seen enough of the path to know the rest of it. A parabola is fixed by very little information. Two numbers in its equation decide the launch, the top and the landing point.
The trajectory with its roots showing
The equation of the trajectory from the last lesson is \(y = x\tan\theta - \dfrac{g x^2}{2u^2\cos^2\theta}\). Both terms contain \(x\), so \(x\) can be taken out as a common factor. The bracket that remains can be written with the range \(R\). The result is called the factored form of the trajectory.
\[ y = x\tan\theta\left(1 - \dfrac{g\,x}{2u^2\cos^2\theta\,\tan\theta}\right) \]
Why. take \(x\tan\theta\) out of both terms of the trajectory equation.
\[ y = x\tan\theta\left(1 - \dfrac{g\,x}{2u^2\sin\theta\cos\theta}\right) \]
Why. \(\cos^2\theta\tan\theta = \cos\theta\sin\theta\).
\[ y = x\tan\theta\left(1 - \dfrac{x}{R}\right) \]
Read it. \(R = \dfrac{2u^2\sin\theta\cos\theta}{g}\) is the range on level ground, so the fraction is \(x/R\).
A product is zero only when one of its factors is zero. The factor \(x\) is zero at the launch point. The factor \(1 - x/R\) is zero at \(x = R\), the landing point. These two values of \(x\) are the roots of the trajectory. A parabola is symmetric about the vertical line through its top, so the top lies exactly halfway between the two roots, at \(x = R/2\). Putting \(x = R/2\) into the factored form gives the maximum height, \(H = \tfrac14 R\tan\theta\).
Read the factored form in two parts. The first factor, \(x\tan\theta\), is the straight line along which the body was launched, the dashed line in Fig 6.1. The second factor, \(1 - x/R\), shrinks that height steadily as \(x\) grows, and reaches zero at the landing point. Gravity is the reason for the second factor.
The trajectory of a projectile on level ground can be written \(y = x\tan\theta\,(1 - x/R)\). The path meets the ground at \(x = 0\) and \(x = R\), and it is symmetric about the vertical line \(x = R/2\). The highest point is at \(x = R/2\), at a height \(H = \tfrac14 R\tan\theta\).
The path of a ball is \(y = x\tan\theta\,(1 - x/R)\). At which values of \(x\) is the ball at ground level?
A ball is thrown at \(45°\) and would land \(40\,\text{m}\) away on level ground. A wall stands \(30\,\text{m}\) from the thrower. How high is the ball when it passes over the wall, in metres?
Reading a launch from its path
An examiner may give the path of a projectile as an equation, such as \(y = \sqrt3\,x - 5x^2\), with \(x\) and \(y\) in metres. Every such path has the form \(y = Ax - Bx^2\). Compare it term by term with the standard trajectory. The coefficient of \(x\) is \(\tan\theta\). The coefficient of \(x^2\) is \(g/(2u^2\cos^2\theta)\), which then gives the launch speed. The range and the height come straight from the roots.
The formula for \(u^2\) uses one identity. Since \(1/\cos^2\theta = 1 + \tan^2\theta = 1 + A^2\), the condition \(B = g/(2u^2\cos^2\theta)\) becomes \(B = g(1 + A^2)/(2u^2)\). Rearranging gives \(u^2\). The range needs no formula at all: \(y = Ax - Bx^2 = x(A - Bx)\) is zero at \(x = A/B\).
A small ball moves along the path \(y = \sqrt3\,x - 5x^2\), with \(x\) and \(y\) in metres. Take \(g = 10\,\text{m/s}^2\). Find the angle of projection, the launch speed, the range and the maximum height.
Compare with \(y = Ax - Bx^2\): \(A = \sqrt3\) and \(B = 5\,\text{m}^{-1}\).
\(\tan\theta = \sqrt3\). \(u^2 = g(1 + A^2)/(2B) = 10 \times 4 / 10\). \(R = A/B\), \(H = A^2/(4B)\).
\(\theta = 60°\). \(u^2 = 4\), so \(u = 2\,\text{m/s}\). \(R = \sqrt3/5 \approx 0.35\,\text{m}\). \(H = 3/20 = 0.15\,\text{m}\).
The path belongs to a gentle toss at \(2\,\text{m/s}\), \(60°\) above the horizontal, like a paper ball flicked towards a dustbin. It rises \(15\,\text{cm}\) and lands \(35\,\text{cm}\) away.
A projectile moves along \(y = x - \dfrac{x^2}{40}\), with \(x\) and \(y\) in metres. What is its range on level ground, in metres?
A projectile moves along \(y = 2x - 0.1x^2\), with \(x\) and \(y\) in metres. Which statements are true?
When the parabola stops being true
The parabola rests on three assumptions. There is no air. The ground is flat. The acceleration \(g\) has the same size and the same direction everywhere along the flight. For a cricket ball or a stone, all three hold well enough. For some motions, one or more of them fails, and the path is no longer a parabola.
Air drag is the push of the air against a moving body, directed opposite to its velocity. It grows quickly with speed. Drag reduces the horizontal velocity, which a parabola keeps constant, and it slows the climb. So the real path is shorter and lower than the parabola. It is also not symmetric: the ball comes down more steeply than it went up. Because of drag, the angle that throws farthest is less than \(45°\).
The Earth is round, with radius \(R_E \approx 6400\,\text{km}\). Over a horizontal distance \(x\), the ground falls below a flat line by about \(x^2/(2R_E)\). Over the \(100\,\text{m}\) of a cricket ground this is less than a millimetre. Over tens of kilometres it is tens of metres. Also, \(g\) always points to the centre of the Earth, so over a very long flight its direction turns. For a flight of thousands of kilometres, \(g\) also becomes weaker with height.
When the direction of \(g\) is allowed to turn towards the centre of the Earth, the path of a body in free flight is not a parabola but part of an ellipse, with the centre of the Earth at one focus. The parabola is an excellent approximation to the top of that ellipse when the flight is short. The chapter on gravitation returns to this.
Newton imagined a cannon on a very high mountain, firing horizontally. A slow shot falls to the ground nearby, along a path close to a parabola. A faster shot lands farther away, and the Earth's surface curves away beneath it as it falls. Newton reasoned that at a great enough speed the ground would curve away as fast as the shot falls. The shot would then never land and would circle the Earth. The same fall that bends a cricket ball into a parabola keeps the Moon in its orbit.
A shell flies \(20\,\text{km}\) horizontally. Roughly how many metres has the Earth's surface dropped below a flat plane through the gun at the landing point? Take \(R_E = 6400\,\text{km}\).
A long-range gun sends a shell \(30\,\text{km}\). Why does the parabola predict its landing point badly?
Explain why the highest point of a projectile on level ground is at exactly half the range, using the factored form of the trajectory.
The factored form \(y = x\tan\theta\,(1 - x/R)\) is zero at \(x = 0\) and at \(x = R\), the launch and landing points. A parabola is symmetric about the vertical line through its top, so two points at the same height lie at equal distances on either side of that line. The launch and landing points are at the same height, so the top is midway between them, at \(x = R/2\). In terms of the motion, the climb and the fall take equal times, and the horizontal velocity is constant, so equal horizontal distances are covered in each half.
In 1918 a German long-range gun shelled Paris from about \(120\,\text{km}\) away. Its shells climbed to about \(40\,\text{km}\), where the thin air offered little drag, and they stayed in the air for about three minutes. At such distances a flat-Earth parabola was useless. The gunners had to allow for the curvature of the Earth and even for the rotation of the Earth beneath the shell during its flight.
- Reading the angle from the coefficient of \(x^2\), or reading \(u\) from the coefficient of \(x\).The coefficient of \(x\) is \(\tan\theta\) alone. The launch speed comes only from the coefficient of \(x^2\), after \(\theta\) is known.
- Using the factored form or \(H = R\tan\theta/4\) for a launch from a height.Both assume landing at the launch level. From a height, return to the two motions and the vertical quadratic.
- Treating the parabola as exact for every flight.It holds when air drag is small and the flight is short enough for the ground to be flat and \(g\) to be fixed.