The shape of the path
The two equations \(x = (u\cos\theta)\,t\) and \(y = (u\sin\theta)\,t - \tfrac12 gt^2\) describe where the projectile is at each time. To find the shape of the path, remove the time. Write \(t\) in terms of \(x\) from the first equation and substitute into the second. The result is called the equation of the trajectory.
\[ t = \dfrac{x}{u\cos\theta} \]
Why. from \(x = (u\cos\theta)\,t\). The horizontal motion gives the time at which the projectile is at horizontal position \(x\).
\[ y = (u\sin\theta)\dfrac{x}{u\cos\theta} - \dfrac{1}{2} g\left(\dfrac{x}{u\cos\theta}\right)^2 \]
Why. substitute this \(t\) into the vertical equation.
\[ y = x\tan\theta - \dfrac{g\,x^2}{2u^2\cos^2\theta} \]
Read it. \(\sin\theta / \cos\theta = \tan\theta\), and the square is written out. This is \(y = Ax - Bx^2\) with constants \(A\) and \(B\), the equation of a parabola.
With air resistance neglected, the path of a projectile is a parabola. The parabola is symmetric about the vertical line through the highest point, so the projectile crosses any height twice, at equal horizontal distances from that line, with equal speeds.
The equation \(y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}\) contains \(x\) and \(x^2\) but no higher power. What does this tell us about the path?
Which angle throws farthest?
The range is \(R = u^2\sin 2\theta / g\). For a fixed speed, \(R\) is largest when \(\sin 2\theta\) is largest. The sine of an angle is never more than 1, and it equals 1 when the angle is \(90°\). So \(2\theta = 90°\), that is, \(\theta = 45°\). The maximum range is \(R_{\max} = u^2 / g\).
Now compare two angles that add to \(90°\), such as \(30°\) and \(60°\). Their doubles are \(60°\) and \(120°\), and \(\sin 60° = \sin 120°\). So the two angles give the same range. They do not give the same flight. The \(60°\) throw goes higher and stays longer in the air; the \(30°\) throw is flat and quick.
For a given launch speed on level ground, the angles \(\theta\) and \(90° - \theta\) give the same range. The larger angle gives the greater maximum height and the longer time of flight. The range is greatest at \(45°\), where it equals \(u^2/g\).
Two projectiles are launched with the same speed, one at \(30°\) and the other at \(60°\). Which statement is correct?
A ball is thrown at \(20\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). What is the greatest horizontal distance it can reach on level ground, in metres?
Thrown from a height
The standard results assume that the projectile lands at the level it was launched from. Many real throws do not. A ball thrown from a rooftop lands on the ground below. The method does not change. Split the motion into two, and write the vertical equation with the correct displacement. The time of flight now comes from a quadratic equation, and the standard results for \(T\) and \(R\) no longer apply.
A ball is thrown horizontally at \(8\,\text{m/s}\) from the top of a building \(20\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). Find the time to reach the ground, the horizontal distance from the building, and the speed on landing.
Origin at the launch point, \(x\) horizontal, \(y\) upward. \(u_x = 8\,\text{m/s}\), \(u_y = 0\), \(a_y = -10\,\text{m/s}^2\). The ground is at \(y = -20\,\text{m}\).
Vertical: \(-20 = 0 - 5t^2\). Horizontal: \(x = 8t\). Landing velocity: \(v_x = 8\), \(v_y = -10t\).
\(t^2 = 4\), \(t = 2\,\text{s}\). \(x = 16\,\text{m}\). \(v_y = -20\,\text{m/s}\), so the speed is \(\sqrt{8^2 + 20^2} = \sqrt{464} \approx 21.5\,\text{m/s}\).
The ball lands after \(2\,\text{s}\), \(16\,\text{m}\) from the base of the building, at about \(21.5\,\text{m/s}\). A ball simply dropped from the roof would also land after \(2\,\text{s}\), at the base.
A ball is thrown horizontally from a cliff. Which quantity decides how long the ball takes to reach the sea below?
Thrown at an angle from a height
Now let the throw from the height make an angle \(\theta\) with the horizontal. Put the origin at the launch point, with \(y\) upward, so the ground is at \(y = -h\). The vertical motion ends when \(y = -h\). The vertical equation then gives a quadratic in \(t\).
\[ -h = (u\sin\theta)\,t - \tfrac12 g t^2 \quad\Longrightarrow\quad \tfrac12 g t^2 - (u\sin\theta)\,t - h = 0 \]
A quadratic has two roots. The positive root is the time of flight. The negative root belongs to no real moment of this throw. It is the instant at which the same parabola, traced backwards, would have crossed the ground level.
Three balls leave the top of a \(40\,\text{m}\) cliff at the same instant, all at \(20\,\text{m/s}\). One is aimed \(30°\) up, one level and one \(30°\) down. Which ball reaches the ground first?
Up, level or down?
Watch the three balls in Fig 5.4. The ball aimed down starts with a downward velocity of \(10\,\text{m/s}\), so it reaches the ground in \(2\,\text{s}\). The level ball starts its fall from rest, vertically, and takes \(2.8\,\text{s}\). The ball aimed up spends time climbing, then falls \(45\,\text{m}\) from its highest point, and takes \(4\,\text{s}\). The distance from the cliff is the horizontal velocity multiplied by this time. The upward ball has a smaller horizontal velocity than the level ball, \(17.3\) against \(20\,\text{m/s}\), but its longer time more than makes up for this.
Look at the landing speeds as well. The vertical component obeys \(v_y^2 = u_y^2 + 2gh\), component by component, while \(v_x = u_x\) stays fixed. Adding the squares gives \(v^2 = u_x^2 + u_y^2 + 2gh = u^2 + 2gh\). This result does not contain the angle. So all three balls land at the same speed, \(\sqrt{400 + 800} \approx 34.6\,\text{m/s}\), in different directions.
For a launch from a height, the time of flight comes from the vertical quadratic, and the standard formulas for \(T\), \(H\) and \(R\) do not apply. Complementary angles no longer share a range; the flatter one goes farther. The landing speed is \(\sqrt{u^2 + 2gh}\) for every angle of projection.
A ball is thrown at \(20\,\text{m/s}\) at \(30°\) above the horizontal from the top of a cliff \(40\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). Find the time of flight, the distance from the base of the cliff where the ball lands, and the landing speed.
Origin at the launch point, \(y\) upward. \(u_x = 20\cos 30° = 10\sqrt3 \approx 17.3\,\text{m/s}\), \(u_y = 20\sin 30° = 10\,\text{m/s}\). The ground is at \(y = -40\,\text{m}\).
Vertical: \(-40 = 10t - 5t^2\), that is, \(t^2 - 2t - 8 = 0\). Horizontal: \(x = 17.3\,t\).
\((t - 4)(t + 2) = 0\), so \(t = 4\,\text{s}\); the root \(t = -2\,\text{s}\) is not part of the flight. \(x = 10\sqrt3 \times 4 = 40\sqrt3 \approx 69.3\,\text{m}\). \(v_y = 10 - 40 = -30\,\text{m/s}\).
The ball lands after \(4\,\text{s}\), about \(69.3\,\text{m}\) from the base, at \(\sqrt{300 + 900} \approx 34.6\,\text{m/s}\). The same speed follows from \(\sqrt{u^2 + 2gh} = \sqrt{1200}\).
A stone is thrown at \(20\,\text{m/s}\) at \(30°\) above the horizontal from the edge of a cliff \(15\,\text{m}\) high. Take \(g = 10\,\text{m/s}^2\). How long does the stone take to reach the ground below, in seconds?
On level ground, throws at \(30°\) and \(60°\) with the same speed land at the same distance. Both throws are now made at that speed from the top of a high cliff. Which lands farther from the base?
Explain, without using the formula, why throwing at a very small angle or a very steep angle both give a short range, so that the best angle lies between.
A very flat throw has a large horizontal velocity but almost no vertical velocity, so it stays in the air for a very short time and lands close by. A very steep throw stays in the air for a long time but has almost no horizontal velocity, so it also lands close by. Range is horizontal velocity multiplied by time in the air, and it is largest when neither factor is small. That balance is reached at \(45°\).
A long jumper leaves the board at about \(20°\), not \(45°\). The reason is that the jumper cannot keep the same launch speed at every angle. Running speed is horizontal, and converting it into a steep take-off costs a great deal of speed. The jumper gets a longer jump from a fast, flat take-off than from a slow, steep one. The \(45°\) rule holds only when the launch speed is the same at every angle, as it is for a thrown ball but not for a running human.
- Using \(R = u^2\sin 2\theta / g\) when the launch and landing levels differ.The standard results assume the same level. From a height, write the vertical equation with the actual displacement and solve for \(t\).
- Taking the speed at the top of the flight as zero when computing the landing speed.The horizontal component survives the whole flight. Combine both components on landing.
- Taking \(u_y\) as positive for a throw aimed below the horizontal.Fix one sign convention before writing any equation. With \(y\) upward, a downward aim has a negative \(u_y\), and the ground is at \(y = -h\).
- Keeping the negative root of the vertical quadratic, or averaging the two roots.Only the positive root is a moment of the flight. The negative root belongs to the parabola traced backwards.
- Assuming that a steeper throw always goes farther.Height rises with the angle, range peaks at \(45°\) and then falls. Complementary angles share a range.