Lesson 1.7 · Unit 1 · HS-PS2-1
Two motions happening at once, neither affecting the other
A thrown ball curves, and a curve looks like a single complicated motion. It is not. It is two simple motions running side by side: constant velocity horizontally and free fall vertically. Separating them turns every projectile problem into two problems already solved in this unit.
- A vector can be split into perpendicular components, with \( v_x = v\cos\theta \) and \( v_y = v\sin\theta \).
- Components recombine by the Pythagorean theorem, \( v = \sqrt{v_x^2 + v_y^2} \), with direction \( \theta = \tan^{-1}\left( \dfrac{v_y}{v_x} \right) \).
- Horizontal and vertical motion are independent. Neither affects the other, and gravity acts on one of them only.
- Horizontally there is no acceleration, so \( v_x \) is constant and \( x = v_x t \).
- Vertically the motion is free fall, so all of lesson 1.6 applies unchanged.
- Time is the quantity the two motions share. It is the bridge between them, and nearly every projectile problem is solved by finding it in one direction and using it in the other.
- At the top of the path the vertical velocity is zero, but the horizontal velocity is not, so the projectile is still moving.
Where students lose marks: using the full launch speed in a vertical equation. Only \( v_y \) belongs in the vertical calculation and only \( v_x \) in the horizontal one. Resolve into components first, and keep the two columns separate on the page.
The problem. A ball is fired horizontally at 15 m/s from the edge of a 20 m cliff. (a) Find the time to land. (b) Find how far from the base it lands. (c) Find its speed and direction on impact. (d) Explain why a ball fired horizontally lands at the same time as one simply dropped.
Step one: separate the motions. Take right positive and up positive, with the origin at the launch point. Horizontal: \( v_x = 15\ \text{m/s} \), \( a_x = 0 \). Vertical: \( v_{0y} = 0 \) because the launch was horizontal, \( a_y = -9.8\ \text{m}/\text{s}^2 \), and the ground is at \( \Delta y = -20\ \text{m} \).
Step two: find the time from the vertical motion for (a). Time is the shared quantity, and the vertical column is the one that determines it, because the ground is at a fixed height rather than a fixed distance. \( -20 = 0 + \tfrac{1}{2}(-9.8)t^2 \) \( t^2 = \dfrac{40}{9.8} = 4.082 \), so \( t = 2.02\ \text{s} \).
Step three: use that time horizontally for (b). \( x = v_x t = 15 \times 2.02 = 30.3\ \text{m} \). This is the whole method. One column gives the time, the other consumes it.
Step four: find the impact velocity components for (c). Horizontal: unchanged at \( v_x = 15\ \text{m/s} \), because nothing accelerated it. Vertical: \( v_y = v_{0y} + a_yt = 0 + (-9.8)(2.02) = -19.8\ \text{m/s} \), the sign meaning downward.
Step five: recombine them. \( v = \sqrt{15^2 + 19.8^2} = \sqrt{225 + 392} = \sqrt{617} = 24.8\ \text{m/s} \). Direction: \( \theta = \tan^{-1}\left( \dfrac{19.8}{15} \right) = 52.9^\circ \) below the horizontal. Check the size. The resultant must exceed both components and be less than their sum, so it must lie between 19.8 and 34.8. It does ✓
Step six: begin (d). Compare two balls released from the same height at the same moment, one dropped and one fired horizontally at 15 m/s. Write the vertical column for each. Dropped: \( v_{0y} = 0 \), \( a_y = -9.8 \). Fired: \( v_{0y} = 0 \), \( a_y = -9.8 \). The two columns are identical. The horizontal velocity does not appear in the vertical equations at all.
Step seven: state why it cannot appear. Acceleration comes from force, and the only force acting is gravity, which points straight down. A downward force changes only the downward velocity. Nothing pushes or resists horizontally, so the horizontal motion proceeds as though gravity were absent and the vertical motion proceeds as though the horizontal motion were absent. Both balls therefore satisfy \( -20 = -4.9t^2 \) and both land at 2.02 s.
Step eight: note the limits and the demonstration. This is exactly right in the absence of air resistance, which is what makes it a physics result rather than a household one. In air, the fired ball moves faster and meets more drag, so it lands very slightly later. The classic demonstration uses a spring-loaded device that drops one ball and fires another horizontally at the same instant. The two strike the floor together, and with a hard floor the single click rather than two is the evidence. What the result buys. Every projectile problem in this course reduces to two independent one-dimensional problems joined only by the shared time, and both of those were solved in lessons 1.5 and 1.6. Nothing new has to be learned to handle a curve.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning or a multi-step calculation. Ignore air resistance.
- What is the horizontal acceleration of a projectile?
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Zero
- What is its vertical acceleration?
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\( 9.8\ \text{m}/\text{s}^2 \) downward
- Which quantity is shared between the horizontal and vertical calculations?
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Time
- A projectile is launched at 20 m/s at \( 30^\circ \). Find its horizontal component.
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\( 20\cos 30^\circ = 20(0.866) \). 17.3 m/s
- Find its vertical component.
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\( 20\sin 30^\circ = 20(0.5) \). 10.0 m/s
- A stone is thrown horizontally at 25 m/s from a 45 m cliff. Find the time to land and
the horizontal distance.
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Vertical: \( 45 = \tfrac{1}{2}(9.8)t^2 \), so \( t^2 = \dfrac{90}{9.8} = 9.184 \) and \( t = 3.03\ \text{s} \). Horizontal: \( x = 25 \times 3.03 = 75.8\ \text{m} \). The launch speed never entered the vertical calculation, which is the point of the method. 3.03 s and 75.8 m
- For the projectile of questions 4 and 5, find the time of flight over level ground.
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It rises and falls symmetrically, so the total flight is twice the time to the top. Time to top: \( \dfrac{v_y}{g} = \dfrac{10.0}{9.8} = 1.02\ \text{s} \). Total: \( 2 \times 1.02 = 2.04\ \text{s} \). 2.04 s
- Find the range and maximum height of that same projectile.
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Range: \( x = v_x t = 17.3 \times 2.04 = 35.3\ \text{m} \). Maximum height: \( \dfrac{v_y^2}{2g} = \dfrac{100}{19.6} = 5.1\ \text{m} \). Check the height a second way. Rising for 1.02 s from 10.0 m/s: \( \dfrac{10.0 + 0}{2}(1.02) = 5.1\ \text{m} \) ✓ Range 35.3 m, maximum height 5.1 m
- Explain why the horizontal and vertical motions of a projectile are independent.
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Because the only force acting is vertical, so it can only change the vertical velocity, and perpendicular directions do not exchange information. Start from the cause. Acceleration is produced by force. For a projectile the only force is gravity, which points straight down. A vertical force changes only the vertical component. It has no horizontal part, so \( a_x = 0 \) and \( v_x \) is unchanged for the whole flight. And nothing acts horizontally, so the horizontal motion has no way to influence the vertical one either. The independence runs both ways. What makes this legitimate mathematically. Perpendicular components of a vector can be treated separately because neither contributes to the other. That is a property of perpendicularity, not a physical assumption, and it is why components are chosen at right angles in the first place. The consequence that surprises people. A bullet fired horizontally and a bullet dropped hit the ground together. The fired one travels hundreds of meters while doing so, and none of that horizontal travel delays its fall by any amount. Where the independence fails. Air resistance acts opposite to the total velocity, so it has both components, and each one then depends on the other. That couples the two motions, and projectile problems with drag cannot be separated this way. It is the idealization that makes the method work. Gravity is purely vertical, so it cannot alter the horizontal motion
- A ball is kicked at 20 m/s at \( 30^\circ \) above the horizontal from the top of a
15 m wall, landing on the ground below. Find the time of flight and the horizontal
distance from the wall.
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Resolve the launch velocity. \( v_x = 20\cos 30^\circ = 17.3\ \text{m/s} \) \( v_{0y} = 20\sin 30^\circ = +10.0\ \text{m/s} \) Set the vertical frame. Up positive, origin at the launch point, so the ground is at \( \Delta y = -15\ \text{m} \) and \( a_y = -9.8\ \text{m}/\text{s}^2 \). Write the vertical equation. \( -15 = 10.0t + \tfrac{1}{2}(-9.8)t^2 \) \( -15 = 10.0t - 4.9t^2 \) \( 4.9t^2 - 10.0t - 15 = 0 \) Solve it. Discriminant: \( 100 - 4(4.9)(-15) = 100 + 294 = 394 \). \( \sqrt{394} = 19.85 \). \( t = \dfrac{10.0 \pm 19.85}{9.8} \). Positive root: \( t = \dfrac{29.85}{9.8} = 3.05\ \text{s} \). The negative root, \( -1.01\ \text{s} \), is rejected as before. Find the horizontal distance. \( x = v_x t = 17.3 \times 3.05 = 52.8\ \text{m} \). Check the vertical result by stages. Rising: the ball takes \( \dfrac{10.0}{9.8} = 1.02\ \text{s} \) to reach the top, climbing \( \dfrac{100}{19.6} = 5.1\ \text{m} \) above the wall, so it is 20.1 m above the ground there. Falling: from rest through 20.1 m takes \( \sqrt{\dfrac{2(20.1)}{9.8}} = \sqrt{4.102} = 2.03\ \text{s} \). Total: \( 1.02 + 2.03 = 3.05\ \text{s} \) ✓ matching the quadratic. That two-stage check is worth the extra minute, because it tests the physics rather than repeating the algebra: if the quadratic had been set up with the wrong sign on the 15, the stages would disagree. Sanity check the distance. Flying for 3.05 s at 17.3 m/s gives about 53 m, and the ball spends longer aloft than it would on level ground because it lands lower than it started. Both are consistent ✓ 3.05 s and 52.8 m from the wall