Senior 4 practice book · Unit 7 of 10

Projectile and Uniform Circular Motion

50 questions that complete the Senior 4 quiz for this unit: 21 core and 29 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A thrown ball keeps moving forward because the force of the throw stays with it.Once the ball leaves the hand, the throwing force stops; the ball keeps its horizontal velocity by inertia and the only force on it (no air resistance) is gravity.
A ball thrown faster horizontally stays in the air longer before landing.The time of fall depends only on the height and g; a faster ball lands farther away, but at the same moment as a slower one.
A body moving in a circle at constant speed has no acceleration.Its velocity keeps changing direction, so it accelerates towards the centre (a = v²/r), and a resultant (centripetal) force must act on it.
In a turning vehicle an outward "centrifugal force" throws passengers outwards.No outward force acts; the passenger tends to keep moving in a straight line, and the door or seat pushes inwards to make the passenger turn.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Nature of projectile motion: gravity only, independent horizontal and vertical motion
Horizontal launch: time of flight and horizontal distance
Velocity of a projectile at any instant: speed and direction
Angled launch from level ground: components, time of flight, maximum height, range
Trajectory equation and height at a given horizontal distance (clearing obstacles)
Angled launch from a height (landing below the launch point)
Real projectiles: effect of air resistance
Describing circular motion: radians, period, frequency, angular velocity, v = ωr
Centripetal acceleration and centripetal force
Level bends (friction) and banked roads
Conical pendulum
Motion in a vertical circle: forces at the top and bottom
Circular motion in daily life: tangent release, centrifuges, spin dryers, satellites

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 29 harder ones are marked ★ Challenge.

1True or false
Air resistance on a projectile always acts vertically upward, opposite to gravity.
Show answer
Answer: False
Drag acts opposite to the velocity, so it points backwards and downwards on the way up and backwards and upwards on the way down.
Common mistakeLearners often treat drag like a falling object's drag; for a projectile the velocity is at an angle, so the drag is too.
2True or false · ★ Challenge
The centripetal force does no work on a body in uniform circular motion, because it is always perpendicular to the velocity.
Show answer
Answer: True
Work = Fs cos 90° = 0, so the kinetic energy (and speed) stays constant.
Common mistakeLearners sometimes think any force must change the speed; a force perpendicular to the motion changes only the direction.
3Multiple choice · ★ Challenge
With air resistance, the launch angle that gives the greatest range on level ground is usually:
1. Aa little less than 45°
2. Bexactly 45°, as on paper
3. Ca little more than 45°
4. Dexactly 90°, straight up
Show answer
Answer: A. a little less than 45°
Drag acts for longer on a high, slow path, so a slightly lower, faster path loses less and goes farther.
Common mistakeChoosing exactly 45° applies the no-air-resistance result to a real ball.
4True or false
A stone thrown horizontally from a cliff hits the ground with a speed greater than its launch speed.
Show answer
Answer: True
It keeps its horizontal speed and gains a vertical component, so the resultant speed is larger.
Common mistakeSome learners think the speed stays equal to the launch speed because the horizontal speed is constant; the vertical part adds to it.
5Multiple choice · ★ Challenge
A satellite moves in a circular orbit round the Earth at constant speed. What provides its centripetal force?
1. AIts rocket engine firing all the time
2. BA balance of gravity and centrifugal force
3. CThe Earth's gravitational pull
4. DNothing, since there is no gravity in space
Show answer
Answer: C. The Earth's gravitational pull
Gravity pulls the satellite towards the Earth's centre and keeps changing the direction of its velocity.
Common mistakeChoosing "no gravity in space" is a common error: at satellite heights gravity is still nearly as strong as on the ground.
6True or false · ★ Challenge
The trajectory equation y = x tan θ − gx²/(2u² cos² θ) contains an x² term, which is why the path of a projectile is a parabola.
Show answer
Answer: True
Height depends on x and x²; a curve y = ax − bx² is a parabola.
Common mistakeSome learners think the path is part of a circle; a circle would need x² and y² terms, not just x².
7True or false
All points on a turning merry-go-round have the same angular velocity, but points nearer the edge have a greater linear speed.
Show answer
Answer: True
Every point turns through the same angle in the same time; v = ωr, so a larger radius gives a larger speed.
Common mistakeSome learners think the outer points also turn "faster" in rad/s; they only cover more distance per turn.
8Multiple choice · ★ Challenge
A centrifuge in a hospital laboratory spins tubes of blood. Why do the denser red blood cells collect at the far end of each tube, away from the axis?
1. AThey need more centripetal force than the liquid around them gives
2. BAn outward force pulls hardest on the densest particles
3. CThe red cells are attracted magnetically to the bottom
4. DThe liquid is pushed towards the axis by the spinning
Show answer
Answer: A. They need more centripetal force than the liquid around them gives
Each part needs a force mv²/r; denser cells need more than the pressure of the surrounding liquid supplies, so they move outwards until the tube end pushes them round.
Common mistakeThe "outward force" option repeats the centrifugal idea; the cells drift outwards because the inward force on them is too small.
9True or false
If air resistance is ignored, a projectile has the same acceleration at every point of its path: about 10 m/s² vertically downward.
Show answer
Answer: True
Only gravity acts, so the acceleration is g downward on the way up, at the top and on the way down.
Common mistakeMany learners think the acceleration changes direction at the top; the velocity changes, but the acceleration stays g downward.
10Multiple choice · ★ Challenge
On a level bend the greatest safe speed is set by friction: v = √(μgr). On a rainy day μ falls from 0.60 to 0.30. How does the greatest safe speed change?
1. AIt becomes half as large
2. BIt becomes 0.71 times as large
3. CIt becomes a quarter as large
4. DIt does not change at all
Show answer
Answer: B. It becomes 0.71 times as large
v ∝ √μ, so the new speed is √(0.30/0.60) = √0.5 ≈ 0.71 of the old one.
Common mistakeChoosing "half" forgets the square root: halving μ halves v², not v.
11True or false
A ball thrown at 30° below the horizontal from a balcony lands sooner than one thrown horizontally at the same speed from the same balcony.
Show answer
Answer: True
The downward throw starts with a downward vertical velocity, so it covers the height in less time.
Common mistakeThinking all throws from the same height take the same time is only true when the initial vertical velocity is the same (zero).
12Multiple choice · ★ Challenge
The bob of a conical pendulum is made to go round faster on the same string. What happens?
1. AThe angle stays the same and only the tension in the string becomes larger
2. BThe string makes a larger angle with the vertical and its tension rises
3. CThe angle becomes smaller and the tension falls
4. DThe bob rises until the string is horizontal
Show answer
Answer: B. The string makes a larger angle with the vertical and its tension rises
tan θ = v²/(rg) grows with speed, and T = mg/cos θ grows as θ grows; the string can never become horizontal because T cos θ must still balance mg.
Common mistakeChoosing "string becomes horizontal" forgets that a vertical component of tension is always needed to hold up the weight.
13True or false
At the lowest point of a vertical circle, the tension in the string is less than the weight of the stone.
Show answer
Answer: False
At the bottom T − mg = mv²/r, so T = mg + mv²/r, which is greater than the weight.
Common mistakeLearners often copy the top-of-circle result; at the bottom the centre is above, so the tension must exceed the weight.
14Multiple choice · ★ Challenge
The grinding disc of a maize mill turns at 1500 revolutions per minute. What is its angular velocity?
1. A25 rad/s
2. B157 rad/s
3. C9420 rad/s
4. D4.0 rad/s
Show answer
Answer: B. 157 rad/s
f = 1500 ÷ 60 = 25 rev/s; ω = 2πf = 2π × 25 ≈ 157 rad/s.
Common mistakeChoosing 9420 rad/s multiplies 1500 by 2π without changing minutes to seconds.
15Multiple choice
How does the spin cycle of a washing machine remove water from wet clothes?
1. AAn outward centrifugal force pushes the water away from the wet clothes
2. BThe spinning heats the water so that it evaporates away
3. CThe drum pushes the clothes round, but water escapes through the holes
4. DThe drum squeezes the clothes flat against the motor
Show answer
Answer: C. The drum pushes the clothes round, but water escapes through the holes
The drum wall supplies the centripetal force on the clothes; at the holes nothing holds the water in a circle, so it carries on along a tangent and leaves.
Common mistakeThe "outward force" option is the centrifugal-force misconception: the water leaves because no inward force acts on it.
16Multiple choice · ★ Challenge
A bucket of water is swung in a vertical circle of radius 1.0 m. What is the least speed at the top so that the water does not fall out? (g = 10 m/s²)
1. A10 m/s
2. B4.5 m/s
3. C1.0 m/s
4. D3.2 m/s
Show answer
Answer: D. 3.2 m/s
At the least speed gravity alone gives the centripetal force: mg = mv²/r, so v = √(gr) = √10 ≈ 3.2 m/s.
Common mistakeChoosing 10 m/s forgets the square root: v² = gr = 10 m²/s².
17Multiple choice · ★ Challenge
A goalkeeper kicks a ball at 30 m/s. Just after the kick its vertical component of velocity is 18 m/s. What is its horizontal component?
1. A12 m/s
2. B48 m/s
3. C35 m/s
4. D24 m/s
Show answer
Answer: D. 24 m/s
u_x = √(30² − 18²) = √(900 − 324) = √576 = 24 m/s.
Common mistakeChoosing 12 m/s subtracts the speeds directly; components combine by Pythagoras, not by simple addition or subtraction.
18Multiple choice
With air resistance, how does the real path of a long football kick differ from the ideal parabola?
1. AHigher and longer, since the air lifts the ball as it moves along
2. BExactly the same, since air has almost no mass at all
3. CLower and shorter, and it falls more steeply than it rose
4. DA straight line, since air resistance cancels gravity
Show answer
Answer: C. Lower and shorter, and it falls more steeply than it rose
Drag takes energy from the ball and slows the horizontal motion, so the path is lower, shorter and not symmetrical.
Common mistakeChoosing "exactly the same" underestimates drag: for a fast, light ball, drag can be as large as its weight.
19Fill in the blank · ★ Challenge
A conical pendulum bob moves in a circle of radius 0.30 m with the string at 45° to the vertical (g = 10 m/s²). Its speed is ______ m/s (2 s.f.).
Show answer
Answer: 1.7
v² = rg tan θ = 0.30 × 10 × 1 = 3.0, so v = √3.0 ≈ 1.7 m/s.
Common mistakeWriting 3.0 forgets to take the square root of v².
20Short answer · ★ Challenge
A flash photograph of a ball launched horizontally shows its positions at equal time intervals. Describe how the horizontal and the vertical spacings of the images change, and explain why.
Show answer
Model answer: Horizontal spacings are all equal, because no horizontal force acts and the horizontal velocity is constant. Vertical spacings grow steadily (in the ratio 1 : 3 : 5 : ...), because the vertical velocity increases uniformly under gravity.
Common mistakeSaying both spacings increase mixes the two motions; gravity only affects the vertical motion.
21Multiple choice
A car goes round a roundabout at a constant speed. Which statement about its acceleration is correct?
1. AIt is directed towards the centre of the roundabout
2. BIt is zero, because the speed is not changing at all
3. CIt is directed along the road, in the direction of motion
4. DIt is directed outwards, away from the centre of the circle
Show answer
Answer: A. It is directed towards the centre of the roundabout
The direction of the velocity keeps changing, which needs an acceleration v²/r towards the centre.
Common mistakeChoosing zero confuses speed with velocity: a change in direction is a change in velocity.
22Multiple choice · ★ Challenge
A stone is thrown horizontally at 12 m/s from a bridge (g = 10 m/s²). What is its speed 0.50 s later?
1. A17 m/s
2. B12 m/s
3. C13 m/s
4. D7.0 m/s
Show answer
Answer: C. 13 m/s
v_y = gt = 10 × 0.50 = 5.0 m/s; v = √(12² + 5.0²) = √169 = 13 m/s.
Common mistakeChoosing 17 m/s adds the components as numbers; they are at right angles, so use Pythagoras.
23Fill in the blank
For water to stay in a bucket at the top of a vertical circle of radius r, the speed there must be at least √(______).
Show answer
Answer: gr
At the least speed the weight alone supplies the centripetal force: mg = mv²/r, so v = √(gr).
Common mistakeWriting √(2gr) or √(g/r) comes from mixing this result with energy formulas or dividing the wrong way.
24Fill in the blank · ★ Challenge
A ball is kicked at 20 m/s at 53° above the horizontal (sin 53° = 0.8, g = 10 m/s²). It reaches its highest point ______ s after the kick.
Show answer
Answer: 1.6
u_y = 20 × 0.8 = 16 m/s; time to the top = u_y/g = 16 ÷ 10 = 1.6 s.
Common mistakeWriting 3.2 s gives the whole time of flight; the top is reached halfway through.
25Multiple choice · ★ Challenge
A 0.50 kg stone on a 0.80 m string is whirled in a vertical circle at a constant 4.0 m/s (g = 10 m/s²). What is the tension at the lowest point?
1. A15 N
2. B10 N
3. C5.0 N
4. D20 N
Show answer
Answer: A. 15 N
mv²/r = 0.50 × 16 ÷ 0.80 = 10 N. At the bottom T − mg = mv²/r, so T = 10 + 5.0 = 15 N.
Common mistakeChoosing 5.0 N gives the tension at the top, where the weight helps; at the bottom the string must also support the weight.
26Fill in the blank
A ball is launched horizontally from a ledge. If it is launched from a ledge four times as high, its time of flight is ______ times as long.
Show answer
Answer: 2
t = √(2h/g), so multiplying h by 4 multiplies t by √4 = 2.
Common mistakeWriting 4 forgets the square root: the time depends on √h, not on h.
27Multiple choice
Through what angle does the second hand of a clock turn in 15 s?
1. Aπ rad
2. B90 rad
3. Cπ/4 rad
4. Dπ/2 rad
Show answer
Answer: D. π/2 rad
One turn (60 s) is 2π rad, so 15 s is a quarter turn: 2π ÷ 4 = π/2 rad (90°).
Common mistakeChoosing 90 rad mixes up degrees and radians: 90° equals π/2 ≈ 1.57 rad.
28Multiple choice · ★ Challenge
The path of a ball kicked from the ground is y = 0.75x − 0.05x², with x and y in metres. How far from the kicking point does it land?
1. A15 m
2. B7.5 m
3. C30 m
4. D0.75 m
Show answer
Answer: A. 15 m
It lands when y = 0: x(0.75 − 0.05x) = 0, so x = 0.75 ÷ 0.05 = 15 m.
Common mistakeChoosing 7.5 m gives the x-position of the highest point, which is halfway to the landing point.
29Short answer · ★ Challenge
Explain why a passenger in a minibus going over a hump-backed bridge feels lighter at the top. At what speed would the minibus just lose contact with the road if the hump has a radius of 40 m? (g = 10 m/s²)
Show answer
Model answer: At the top, mg − N = mv²/r, so the normal force N = mg − mv²/r is less than the weight and the passenger feels lighter. Contact is lost when N = 0: v = √(gr) = √(10 × 40) = 20 m/s (72 km/h).
Common mistakeAdding mv²/r to mg (as at the bottom of a dip) would make the passenger heavier; at a hump the centre of the circle is below the passenger.
30Multiple choice
A ball is kicked at 50° above the horizontal from level ground. Ignoring air resistance, which is the same at launch and just before landing?
1. AIts vertical velocity, which points upward at both moments
2. BIts speed, though it now moves at 50° below the horizontal
3. CIts direction of motion, which stays at 50° above the horizontal
4. DNothing, as gravity has changed every part of its motion
Show answer
Answer: B. Its speed, though it now moves at 50° below the horizontal
The horizontal component is unchanged and the vertical component has the same size but opposite direction, so the speed is the same.
Common mistakeChoosing the vertical velocity ignores direction: it is upward at launch and downward at landing.
31Fill in the blank · ★ Challenge
A ball leaves a table horizontally at 3.0 m/s and hits the floor 0.40 s later (g = 10 m/s²). Its speed just before landing is ______ m/s.
Show answer
Answer: 5.0
v_y = 10 × 0.40 = 4.0 m/s; v = √(3.0² + 4.0²) = 5.0 m/s.
Common mistakeWriting 4.0 m/s gives only the vertical component; the horizontal 3.0 m/s is still there at landing.
32Multiple choice
Why does a cyclist lean inwards when riding round a bend?
1. ATo make the weight of the bicycle pull the cyclist outwards
2. BTo reduce the friction that is needed between the tyres and the road surface
3. CTo slow the bicycle down by pressing harder on the brakes
4. DSo the road's push gives a sideways force to the centre without toppling
Show answer
Answer: D. So the road's push gives a sideways force to the centre without toppling
Friction at the tyres provides the centripetal force; leaning makes the combined road force pass through the centre of mass, so there is no turning effect that would tip the cyclist over.
Common mistakeChoosing "reduce friction" is wrong on a level road: friction is what supplies the centripetal force, so it is still needed.
33Short answer · ★ Challenge
Explain why R = u² sin 2θ/g predicts well the range of a steel ball thrown 5 m in a classroom, but badly the range of a table-tennis ball or of a long golf shot.
Show answer
Model answer: For a small, heavy steel ball moving slowly, air resistance is tiny compared with its weight, so the ideal formula fits. A table-tennis ball is very light, so drag is large compared with its weight; a golf ball moves very fast (drag grows with speed) and its spin gives lift. In both cases the real range differs a lot from the formula.
Common mistakeSaying "the formula only works for heavy objects" misses the speed effect: even a heavy ball is affected if it moves fast enough.
34Multiple choice · ★ Challenge
A learner measures the force needed to keep a 0.20 kg mass moving in a circle of radius 0.50 m at different speeds. A graph of F against v² is a straight line through the origin. What is its gradient?
1. A0.10 kg m
2. B2.5 m/kg
3. C0.40 kg/m
4. D0.20 kg
Show answer
Answer: C. 0.40 kg/m
F = (m/r)v², so the gradient is m/r = 0.20 ÷ 0.50 = 0.40 kg/m.
Common mistakeChoosing 0.10 kg m multiplies m by r; in F = mv²/r the radius divides.
35Fill in the blank
A point 0.30 m from the axle of a wheel turning at 8.0 rad/s moves with a linear speed of ______ m/s.
Show answer
Answer: 2.4
v = ωr = 8.0 × 0.30 = 2.4 m/s.
Common mistakeDividing instead (8.0 ÷ 0.30) gives 27 m/s; the speed grows with the radius, so multiply.
36Multiple choice · ★ Challenge
A volleyball is served at 10 m/s at 37° above the horizontal (sin 37° = 0.6, cos 37° = 0.8, g = 10 m/s²). How much higher than the serving point is it when it has travelled 4.0 m horizontally?
1. A3.0 m
2. B4.25 m
3. C1.25 m
4. D1.75 m
Show answer
Answer: D. 1.75 m
u_x = 8.0 m/s, so t = 4.0 ÷ 8.0 = 0.50 s; y = 6.0 × 0.50 − ½ × 10 × 0.50² = 3.0 − 1.25 = 1.75 m.
Common mistakeChoosing 3.0 m uses y = u_y t and forgets that gravity pulls the ball down by ½gt² during that time.
37Multiple choice
A ball is thrown horizontally from a high window. As it falls, what happens to the angle its velocity makes with the horizontal?
1. AIt stays the same, as the horizontal speed is constant
2. BIt increases steadily but never quite reaches 90°
3. CIt decreases, as the ball moves farther from the wall
4. DIt becomes 90° once the ball has fallen half the height
Show answer
Answer: B. It increases steadily but never quite reaches 90°
tan α = v_y/v_x; v_y grows while v_x is constant, so the path gets steeper, but v_x never becomes zero.
Common mistakeChoosing "stays the same" looks only at v_x; the direction depends on the ratio of both components.
38Multiple choice · ★ Challenge
Water leaves a horizontal hose held 0.80 m above a garden and lands 3.2 m away (g = 10 m/s²). The hose is raised to 1.8 m with the water leaving at the same speed. How far does the water now reach?
1. A4.8 m
2. B7.2 m
3. C3.2 m
4. D2.1 m
Show answer
Answer: A. 4.8 m
t₁ = √(2 × 0.80/10) = 0.40 s, so u = 3.2 ÷ 0.40 = 8.0 m/s. t₂ = √(2 × 1.8/10) = 0.60 s, so x = 8.0 × 0.60 = 4.8 m.
Common mistakeChoosing 7.2 m assumes the range is proportional to the height; the time of fall goes as √h, so the range rises by √(1.8/0.8) = 1.5 times.
39Short answer · ★ Challenge
A parcel leaves a plane horizontally at 50 m/s. Find the size and direction of its velocity 5.0 s later (g = 10 m/s², no air resistance).
Show answer
Model answer: v_x = 50 m/s, v_y = 10 × 5.0 = 50 m/s. v = √(50² + 50²) ≈ 71 m/s. tan α = 50/50 = 1, so α = 45° below the horizontal.
Common mistakeGiving only "71 m/s" is incomplete: velocity is a vector, so its direction must be stated as well.
40Short answer
In the hammer throw, an athlete whirls the hammer round in a circle and then lets go. The landing area is straight ahead of her. At which point of the circle should she release the hammer? Explain.
Show answer
Model answer: At the point where the hammer's velocity (the tangent to the circle) points towards the landing area, which is a quarter-turn before the hammer is straight in front of her. On release the pull of the wire stops, so the hammer moves off along the tangent, not along the radius.
Common mistakeReleasing when the hammer is straight in front assumes it flies outward along the radius; it flies off along the tangent.
41Short answer · ★ Challenge
A 60 kg learner stands 2.0 m from the centre of a merry-go-round that turns once every 4.0 s. Find the centripetal force needed and say what provides it.
Show answer
Model answer: ω = 2π/T = 2π ÷ 4.0 = 1.57 rad/s. a = ω²r = 1.57² × 2.0 ≈ 4.9 m/s². F = ma = 60 × 4.9 ≈ 300 N towards the centre, provided by friction from the floor on the shoes and by the pull of the hand rail.
Common mistakeUsing v = 2πr/T but then forgetting to square v gives a force far too small; keep track of v² in F = mv²/r.
42Short answer
On a swing ride at a fair, chairs hang on chains from a spinning top. Explain why an empty chair and a chair with an adult in it hang out at the same angle.
Show answer
Model answer: Each chair is a conical pendulum: T cos θ = mg and T sin θ = mv²/r give tan θ = v²/(rg). The mass cancels, so at the same speed and radius both chairs hang at the same angle.
Common mistakeSaying the heavier chair hangs lower assumes weight pulls it back; the extra weight is matched by an equally larger tension.
43Short answer · ★ Challenge
A helicopter flying horizontally at 40 m/s and 180 m above the ground drops a bag of supplies to a village. How far before reaching the village (measured horizontally) should the pilot release it? (g = 10 m/s²)
Show answer
Model answer: Time of fall: t = √(2h/g) = √(2 × 180 ÷ 10) = √36 = 6.0 s. Horizontal distance = 40 × 6.0 = 240 m before the village.
Common mistakeReleasing the bag directly above the village ignores its horizontal velocity; it would land 240 m beyond.
44Multiple choice
A plane flying horizontally at constant speed drops a food parcel. Ignoring air resistance, where is the parcel compared with the plane while it falls?
1. ABehind the plane, as the parcel slows down as soon as it is released
2. BDirectly below the plane, as both keep the same horizontal velocity
3. CIn front of the plane, as gravity speeds the parcel up forwards
4. DStraight below the release point on the ground, falling vertically
Show answer
Answer: B. Directly below the plane, as both keep the same horizontal velocity
After release the parcel keeps the plane's horizontal velocity, so it stays under the plane while it falls.
Common mistakeChoosing "behind the plane" assumes something slows the parcel horizontally; with no air resistance nothing does.
45Short answer · ★ Challenge
The Earth turns once on its axis in 24 h. Kigali is very close to the equator, about 6.4 × 10⁶ m from the Earth's axis. Find the angular velocity of the Earth and the speed at which Kigali moves around the axis.
Show answer
Model answer: T = 24 × 3600 = 86 400 s. ω = 2π/T = 2π ÷ 86 400 ≈ 7.3 × 10⁻⁵ rad/s. v = ωr = 7.27 × 10⁻⁵ × 6.4 × 10⁶ ≈ 465 m/s (about 1700 km/h).
Common mistakeLeaving T in hours gives ω in rad/h; convert to seconds before using v = ωr in m/s.
46Short answer · ★ Challenge
A boy standing on a rock 8.0 m above Lake Kivu throws a stone at 10 m/s at 37° above the horizontal (sin 37° = 0.6, cos 37° = 0.8, g = 10 m/s²). Find how long the stone is in the air and how far out from the rock it hits the water.
Show answer
Model answer: u_y = 6.0 m/s up, u_x = 8.0 m/s. Taking up as positive: −8.0 = 6.0t − 5t², so 5t² − 6t − 8 = 0, giving t = (6 + √(36 + 160))/10 = 2.0 s. Distance = 8.0 × 2.0 = 16 m.
Common mistakeUsing T = 2u sin θ/g = 1.2 s ignores the extra 8.0 m fall; the displacement at landing is −8.0 m, not zero.
47Short answer
Without calculating, explain why a high lob kicked at 70° stays in the air longer than a low drive kicked at 20° with the same speed, although both land the same distance away.
Show answer
Model answer: The 70° kick has a larger vertical component (u sin 70°), so it rises higher and stays up longer (T = 2u sin θ/g). But it has a smaller horizontal component; the longer time and smaller horizontal speed give the same range, since sin 140° = sin 40°.
Common mistakeSaying "the lob goes farther because it is in the air longer" forgets that its horizontal speed is much smaller.
48Short answer · ★ Challenge
A motorcycle and rider of total mass 150 kg ride round a level bend of radius 20 m. The coefficient of friction between tyres and road is 0.50 (g = 10 m/s²). Find the greatest speed at which they can take the bend without skidding.
Show answer
Model answer: Friction provides the centripetal force: μmg = mv²/r, so v = √(μgr) = √(0.50 × 10 × 20) = √100 = 10 m/s (36 km/h). The mass cancels.
Common mistakeMultiplying by the mass on only one side gives a wrong answer; the mass appears in both the friction and the centripetal force, so it cancels.
49Multiple choice
A ball is thrown at 30° above the horizontal from the top of a 20 m cliff. Compared with the same throw on level ground, its horizontal distance travelled is:
1. Asmaller, because the cliff takes away part of its speed
2. Bthe same, because the range depends only on speed and angle
3. Cgreater, because it has extra time to fall the 20 m
4. Dzero, because it falls straight down the cliff face
Show answer
Answer: C. greater, because it has extra time to fall the 20 m
The horizontal velocity is the same, but the ball stays in the air longer while it falls the extra 20 m, so it goes farther.
Common mistakeChoosing "the same" applies R = u² sin 2θ/g, which is only true when launch and landing are at the same level.
50Short answer · ★ Challenge
A football is kicked from the ground at 15 m/s at 45°. A wall 2.5 m high stands 10 m away. Using y = x tan θ − gx²/(2u² cos² θ) with g = 10 m/s², decide whether the ball clears the wall.
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Model answer: y = 10 × tan 45° − 10 × 10²/(2 × 15² × cos² 45°) = 10 − 1000/(2 × 225 × 0.5) = 10 − 4.4 = 5.6 m. Since 5.6 m > 2.5 m, the ball clears the wall by about 3 m.
Common mistakeForgetting to square cos θ (or u) in the second term gives a wrong height; check that the second term has units of metres.