Lesson 1.8 · Unit 1 · CED topic 1.5 (two-dimensional motion)
Projectile motion
A basketball arcing toward the hoop, water from a fountain, a stone skipped off a cliff: after the throw, the only force on each is gravity (when air resistance is small), and each follows the same curved path. The path is a parabola, and the reason it is a parabola is the most useful result of Unit 1: the horizontal and vertical motions are independent of each other. The vertical motion is free fall that you already understand, and the horizontal motion is the simplest motion of all, a constant velocity. This lesson puts the two together.
Describe the motion of an object moving in two dimensions.
- Motion in two dimensions can be analyzed using one-dimensional kinematic relationships if the motion is separated into components.
- Projectile motion is a special case of two-dimensional motion that has zero acceleration in one dimension and constant, nonzero acceleration in the second dimension.
Two independent motions
A projectile is an object moving under the influence of gravity alone, after it has been launched. With up taken as positive and air resistance neglected:
- Horizontal: \(a_x = 0\), so \(v_x\) is constant and \(x = x_0 + v_{x0}t\).
- Vertical: \(a_y = -g = -10\ \text{m/s}^2\), so \(v_y = v_{y0} - gt\) and \(y = y_0 + v_{y0}t - \tfrac12 g t^2\).
The only thing the two motions share is the time. Time is the link: you find it from the vertical motion, which has the constant acceleration, and then use it in the horizontal motion to find how far the object has gone sideways. Figure 1.8a shows the result for a launch at 25 m/s: the horizontal arrow is the same length at every position, while the vertical arrow shrinks, vanishes at the peak and reverses.
A key consequence: the sideways motion does not change how fast the object falls. A ball rolled off a table and a ball dropped from the same height at the same moment hit the floor together, as Figure 1.8b shows with a cliff. The launched ball covers a great deal of ground sideways, but its vertical position at each instant matches the dropped ball’s exactly.
The launch at an angle
For a launch at speed \(v_0\) at an angle \(\theta\) above the horizontal, resolve the initial velocity: \(v_{x0} = v_0\cos\theta\) and \(v_{y0} = v_0\sin\theta\). Then:
- The time to the peak is when \(v_y = 0\): \(t_{\text{up}} = v_{y0}/g\).
- The maximum height above the launch point is \(H = v_{y0}^2/(2g)\).
- For a landing at the launch height, the time of flight is twice the time to the peak, \(T = 2v_{y0}/g\), and the range is \(R = v_{x0}T\).
- At the peak the velocity is purely horizontal, \(v = v_{x0}\), the smallest speed of the flight. The acceleration there is still \(10\ \text{m/s}^2\) downward.
For the same launch speed, the range is greatest at 45\(^\circ\) on level ground, and two angles that add up to 90\(^\circ\) have equal ranges, because \(R = (v_0^2/g)\sin 2\theta\). The steeper of the two paths rises higher and stays in the air longer (Figure 1.8c). These are results to remember, but the exam is more interested in your ability to explain them with the independence of the components than in your ability to recall them.
Component graphs complete the picture. The horizontal position \(x\) grows along a straight line and the horizontal velocity \(v_x\) is a flat line. The vertical velocity \(v_y\) falls along a straight line of slope \(-10\ \text{m/s}^2\) and the vertical position \(y\) is a downward-opening parabola. The vertical graphs are exactly the free-fall graphs of lesson 1.5.
A ball is thrown horizontally at 15 m/s from the top of a 45 m cliff. How long is it in the air, how far from the base does it land, and what are the speed and direction of its velocity at impact?
Time. The vertical motion starts from rest vertically. With down positive, \(45 = \tfrac12(10)t^2\), so \(t = \sqrt{9.0} = 3.0\ \text{s}\).
Distance. \(x = v_x t = (15)(3.0) = 45\ \text{m}\).
Impact velocity. The horizontal component is still \(v_x = 15\ \text{m/s}\). The vertical component is \(v_y = gt = 30\ \text{m/s}\) downward. The speed is \(\sqrt{15^2 + 30^2} = \sqrt{1125} = 34\ \text{m/s}\), directed at \(\tan^{-1}(30/15) = 63^\circ\) below the horizontal.
A ball is launched from level ground with a horizontal velocity component of 20 m/s and a vertical component of 15 m/s (a launch speed of 25 m/s at about 37\(^\circ\)). Find the time of flight, the maximum height and the range.
Peak. \(t_{\text{up}} = 15/10 = 1.5\ \text{s}\), so the time of flight is \(T = 3.0\ \text{s}\).
Height. \(H = \dfrac{15^2}{2(10)} = \dfrac{225}{20} = 11\ \text{m}\) (11.25 m, kept as 11 m to two significant figures).
Range. \(R = (20)(3.0) = 60\ \text{m}\).
At the peak, the velocity is 20 m/s horizontal and the acceleration is \(10\ \text{m/s}^2\) down. The ball lands with the same speed it had at launch, 25 m/s, at 37\(^\circ\) below the horizontal.
A ball is thrown from the edge of a 15 m high roof with a horizontal component of 12 m/s and an upward component of 10 m/s. How far from the building does it land, and what is its speed on landing?
Time. The vertical motion is the cliff problem of lesson 1.5: \(-15 = 10t - 5t^2\), so \(t^2 - 2t - 3 = 0\) and \(t = 3.0\ \text{s}\).
Distance. \(x = (12)(3.0) = 36\ \text{m}\).
Landing velocity. \(v_x = 12\ \text{m/s}\), \(v_y = 10 - 10(3.0) = -20\ \text{m/s}\). The speed is \(\sqrt{12^2 + 20^2} = \sqrt{544} = 23\ \text{m/s}\).
- Separate, solve, recombine. Write the vertical equations and the horizontal equations as two columns, with time as the common variable.
- Horizontal acceleration is zero (no air resistance). A question that asks for the net acceleration at the top of a flight has the answer \(g\) downward, not zero.
- Qualitative claims need a reason. To argue that two balls land together, say that they have the same vertical acceleration and the same initial vertical velocity, so the same vertical motion.
- Speed versus components. The speed at any moment is \(\sqrt{v_x^2 + v_y^2}\); the smallest speed of a symmetric flight occurs at the top, where \(v_y = 0\).
- Experiment design. A common free-response task is to predict the landing spot of a ball launched horizontally from a known height: measure the height and the launch speed, compute the time from the height, then the range.
- A marble rolls off a table 1.25 m high at 2.0 m/s. How long does it fall and how far from the table does it land?
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\(t = \sqrt{2(1.25)/10} = 0.50\ \text{s}\). \(x = (2.0)(0.50) = 1.0\ \text{m}\).
- A soccer ball is kicked from level ground at 20 m/s at 30\(^\circ\) above the horizontal. Find the time of flight, the maximum height and the range.
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\(v_{x0} = 20\cos 30^\circ = 17.3\ \text{m/s}\), \(v_{y0} = 20\sin 30^\circ = 10\ \text{m/s}\). \(T = 2(10)/10 = 2.0\ \text{s}\). \(H = 10^2/20 = 5.0\ \text{m}\). \(R = (17.3)(2.0) = 35\ \text{m}\).
- A package is dropped from a plane flying horizontally at 80 m/s at an altitude of 500 m. Ignoring air resistance, how long does it fall, how far ahead of the drop point does it land, and where is the plane at that moment?
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\(t = \sqrt{2(500)/10} = 10\ \text{s}\). \(x = (80)(10) = 800\ \text{m}\) ahead of the release point. The package keeps the plane’s horizontal velocity, so if the plane continues unchanged it is directly above the package the whole time, 500 m above it when the package lands.
- A projectile has horizontal velocity 12 m/s and initial vertical velocity 16 m/s. What are its speed and acceleration at the top of its path?
Show answer
At the top \(v_y = 0\), so the speed is just the horizontal component, 12 m/s. The acceleration is \(10\ \text{m/s}^2\) downward, as at every other point.
- Two balls are launched at the same speed from level ground, one at 40\(^\circ\) and one at 50\(^\circ\). Which stays in the air longer and which goes farther? Explain.
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The 50\(^\circ\) ball has the larger vertical component, so it takes longer to rise and fall and stays in the air longer. The ranges are equal, because 40\(^\circ\) and 50\(^\circ\) add to 90\(^\circ\): the 40\(^\circ\) ball has the larger horizontal speed but less time, and the 50\(^\circ\) ball has the reverse, and the products are the same.
- Sketch in words the four graphs, \(x\)–\(t\), \(y\)–\(t\), \(v_x\)–\(t\) and \(v_y\)–\(t\), for a ball thrown at an angle from level ground and landing at the same height.
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\(x\)–\(t\): a straight line with constant positive slope. \(y\)–\(t\): a downward-opening parabola starting and ending at zero. \(v_x\)–\(t\): a horizontal line at the launch value. \(v_y\)–\(t\): a straight line with slope \(-10\ \text{m/s}^2\), starting at the launch value, crossing zero at the peak and ending at the negative of the launch value.