Parabolas in flight
Tonight's idea
Throw a ball with speed \(v\) at angle \(\theta\) above the horizontal. Gravity pulls down with constant acceleration \(g \approx 9.81\) m/s², nothing pulls sideways (ignore air), and the two directions can be treated separately.
Sideways: constant speed \(v\cos\theta\), so \(x = v\cos\theta \cdot t\). Up and down: starting speed \(v\sin\theta\), acceleration \(-g\), so \(y = v\sin\theta \cdot t - \tfrac12 g t^2\).
Eliminate the time: \(t = x/(v\cos\theta)\), and substituting,
\[y = x\tan\theta - \frac{g}{2v^2\cos^2\theta}\,x^2.\] A quadratic in \(x\) with a negative leading coefficient: an upside-down parabola. Every thrown or kicked object follows one, air aside.
Landing on level ground means \(y = 0\), which gives \(x = \dfrac{2v^2\sin\theta\cos\theta}{g}\), and with \(2\sin\theta\cos\theta = \sin 2\theta\):
Range \(R = \dfrac{v^2 \sin 2\theta}{g}\), biggest at \(\theta = 45°\). Since \(\sin 2\theta = \sin(180° - 2\theta)\), the angles \(\theta\) and \(90° - \theta\) give the same range.
For a shot put the best angle on paper is about 42°, not 45°, because the shot starts two metres up. Real throwers go lower still, 30° to 40°, since arms push harder at flatter angles. The pure parabola is the first approximation; the corrections are the engineering.
Watch one: 20 m/s at 30°
Range: \(R = \dfrac{20^2 \sin 60°}{9.81} = \dfrac{400 \times 0.866}{9.81} = 35.3\) m.
Maximum height: the vertical speed starts at \(v\sin\theta = 10\) m/s and gravity takes \(g\) off it each second; (final speed)² = (initial speed)² \(- 2gh\) reaches zero at \(h = \dfrac{(v\sin\theta)^2}{2g} = \dfrac{100}{19.62} = 5.10\) m.
Time of flight: up and down are symmetric, so \(t = \dfrac{2v\sin\theta}{g} = \dfrac{20}{9.81} = 2.04\) s. Check: \(x = 17.32 \times 2.04 = 35.3\) m.
Your turn
- A football is kicked at 25 m/s at 40° above the horizontal. Find the range, the maximum height and the time of flight. Take \(g = 9.81\) m/s².
- The same ball is kicked at 25 m/s at 50°. Without recalculating, state the range and explain why. Which kick stays in the air longer, and by how much?
- A hose held 1.2 m above the ground squirts water horizontally at 8 m/s. How long does the water take to reach the ground, and how far away does it land? (Horizontal launch: \(\theta = 0\), so the vertical motion is a pure drop.)
- On the Moon, \(g = 1.62\) m/s². How far would the kick in question 1 travel there? Explain from the range formula why the range is multiplied by exactly the factor \(g\) is divided by.
Stretch: as high as it is long
For which launch angle does a ball rise exactly as high as it travels? Set the maximum height \(\dfrac{v^2\sin^2\theta}{2g}\) equal to the range, show the condition is \(\tan\theta = 4\), and find \(\theta\). Why does the answer not depend on \(v\) or \(g\)?
Answers
- \(R = 625 \sin 80° / 9.81 = 62.7\) m. Height \(= (25\sin 40°)^2/(2 \times 9.81) = 13.2\) m. Time \(= 2 \times 25 \sin 40° / 9.81 = 3.28\) s.
- Same range, 62.7 m, because \(\sin 100° = \sin 80°\): complementary angles share a range. The 50° kick is in the air longer: \(2 \times 25\sin 50°/9.81 = 3.90\) s against 3.28 s, about 0.63 s more, because more of its speed is vertical.
- Drop time from \(1.2 = \tfrac12 g t^2\): \(t = \sqrt{2.4/9.81} = 0.495\) s. Distance \(= 8 \times 0.495 = 3.96\) m.
- \(R = 62.7 \times 9.81/1.62 = 380\) m. \(R = v^2\sin 2\theta/g\) has \(g\) in the denominator and nowhere else, so dividing \(g\) by 6.06 multiplies \(R\) by 6.06.
Stretch. \(\dfrac{v^2\sin^2\theta}{2g} = \dfrac{2v^2\sin\theta\cos\theta}{g}\). Cancel \(v^2/g\) and one \(\sin\theta\): \(\sin\theta = 4\cos\theta\), so \(\tan\theta = 4\) and \(\theta \approx 76.0°\). Both height and range scale as \(v^2/g\), so the ratio between them depends only on the angle.