1.5 Vectors and Motion in Two Dimensions — Worked Solutions
Work through Topic 1.5
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A horizontally launched ball has negligible air resistance. Which quantity stays constant during flight?
- Horizontal velocity
- Vertical velocity
- Speed
Step 1. After horizontal release, gravity is the only force in the negligible-drag model.
Why this step? The launch contact has ended; no continuing horizontal push is included in projectile motion.
Step 2. Therefore aₓ = 0 and vₓ stays equal to its launch value.
Why this step? Only a horizontal velocity change would require horizontal acceleration; downward gravity has no x component.
Step 3. Vertical velocity changes and the speed generally grows during the fall.
Why this step? Gravity changes vᵧ, so the full speed √(vₓ² + vᵧ²) is not held fixed by the constant horizontal component.
Back to this question · Back to the lesson
Question 2
A projectile is at the highest point of an angled flight. Which statement is correct?
- Its total velocity is necessarily zero
- Its vertical velocity is zero and acceleration is downward
- Gravity is briefly zero
Step 1. At the peak, vᵧ = 0 because vertical motion changes from upward to downward.
Why this step? The highest point is a vertical turning point, not necessarily a turning point in the horizontal direction.
Step 2. The nonzero horizontal component continues while acceleration remains downward at g.
Why this step? No horizontal force changes vₓ, and gravity persists at the peak just as it does elsewhere in the flight.
Step 3. Choose vertical velocity zero with downward acceleration.
Why this step? Zero total velocity would ignore the angled launch’s horizontal component; zero gravity would prevent the downward turn.
Back to this question · Back to the lesson
Question 3
Horizontal launch speed doubles at fixed table height. How do fall time and range change?
- Time doubles; range quadruples
- Time is unchanged; range doubles
- Both double
Step 1. For horizontal launch from height h, the fall time satisfies h = ½gt², so t = √(2h/g).
Why this step? Both trials have zero initial vertical velocity and the same height and gravity, so changing horizontal speed does not enter the time equation.
Step 2. Range is R = u t; replacing u by 2u gives R_new = 2R at the same t.
Why this step? Horizontal acceleration is zero, and the common flight time multiplies the constant horizontal launch velocity.
Step 3. The time is unchanged and the range doubles.
Why this step? Doubling flight time would require changing the vertical motion, which the question explicitly holds fixed.
Back to this question · Back to the lesson
Question 4
Resolve a 10 m/s launch at 37° using sin 37° ≈ 0.60 and cos 37° ≈ 0.80. Find the speed at the peak and the time to reach it using g = 10 m/s².
Step 1. Resolve v₀ₓ = 10 cos 37° = 8 m/s and v₀ᵧ = 10 sin 37° = 6 m/s.
Why this step? The angle is measured above horizontal, so the adjacent component uses cosine and the upward component uses sine.
Step 2. At the peak, vᵧ = 0 while vₓ = 8 m/s; speed is therefore 8 m/s.
Why this step? Gravity removes the upward component but no horizontal acceleration removes the 8 m/s component.
Step 3. Use 0 = 6 − 10t to find t = 0.60 s.
Why this step? With upward positive, gravity is −10 m/s²; the peak occurs when that change cancels the initial +6 m/s.
Back to this question · Back to the lesson
Question 5
Draw x–time, y–time, vₓ–time, and vᵧ–time graphs for a horizontal launch. Describe one feature that connects each position graph to its velocity graph.
Step 1. Draw x–t as a rising straight line and vₓ–t as a positive horizontal line.
Why this step? A horizontal launch with negligible drag has one constant positive horizontal velocity, equal to the slope of x–t.
Step 2. Draw y–t starting with zero slope and bending downward; draw vᵧ–t as a line falling from zero with slope −g.
Why this step? With upward positive and no initial vertical motion, constant downward acceleration makes vertical velocity increasingly negative.
Step 3. Match each position graph’s local slope to the corresponding velocity graph’s height.
Why this step? That check catches a mistaken straight y–t graph or changing vₓ graph, neither of which fits these component accelerations.
Back to this question · Back to the lesson
Question 6
Derive how horizontal range depends on table height h for fixed horizontal launch speed u. Which axes would linearize experimental range data?
Step 1. For horizontal launch, solve h = ½gt² to get t = √(2h/g).
Why this step? The fall begins with zero vertical velocity; only the vertical height and gravity determine when the floor is reached.
Step 2. Substitute into R = ut: R = u√(2h/g).
Why this step? The same time must describe both components; u remains constant because horizontal acceleration is zero.
Step 3. Plot R against √h, or square the equation and plot R² against h with slope 2u²/g.
Why this step? Each transformed equation is linear when u and g are held fixed, so the slope can test that dependence rather than assuming R is proportional to h.
Step 4. For R² versus h, slope units are m²/m = m.
Why this step? The axis-unit ratio must match 2u²/g, whose units are (m²/s²)/(m/s²) = m; a dimensionless slope would indicate mislabeled axes.