Why is the path of a projectile a parabola?

Eliminating time between the uniform horizontal and accelerating vertical equations leaves a quadratic relation between y and x.

When can energy replace component work?

For speeds and heights only, since gravity is conservative; directions and horizontal positions still need the components.

S4 Physics projectile motion paper 3: landings, walls and energy

2026-11-10

Tutor Prep Guide · S4 Physics Projectile Motion

Ten harder cases: launching to a lower level, finding the height at a wall and checking landing speed with energy.

Direct answers　Items 01 to 03 cover launching to a lower level, 04 to 06 energy checks and equal landing times, 07 to 09 table edges, bullets and wall heights, and 10 confusing zero velocity with zero acceleration.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Physics Projectile Motion
Level	S4

02　Questions (10 items)

FoundationA ball leaves a 20 m platform at 15 m/s at 30 degrees above the horizontal. Find the time to reach the ground.
FoundationHow far horizontally does that ball travel before landing?
FoundationFind the vertical speed of the same ball when it lands.
IntermediateFind the landing speed of that ball and check it with energy.
IntermediateA ball is launched at 30 m/s at 45 degrees on level ground. Find the range.
IntermediateTwo balls start at the same height, one dropped and one thrown horizontally. Which lands first?
ChallengeA ball leaves a 1.8 m table horizontally at 2.5 m/s. Find the horizontal distance from the table edge.
ChallengeA bullet leaves a gun horizontally at 500 m/s from 1.5 m above the ground. Find the horizontal distance travelled.
ChallengeA ball is launched at 20 m/s at 60 degrees towards a 10 m wall 30 m away. Find its height at the wall.
Error redoA student says a horizontally thrown ball does not accelerate downwards and so lands later. Give the correct conclusion.

03　Answers and one-line notes

#	Answer	One-line note
1	1.39 s	With up positive, -20 = 7.5 t – 4.905 t squared, so t = 1.39 s; the drop must carry a minus sign.
2	18.1 m	The horizontal speed is a constant 12.99 m/s, so x = 12.99 x 1.39 = 18.1 m.
3	21.2 m/s downwards	v = 7.5 + 9.81 x 1.39 = 21.2 m/s; the sign gives the direction and the size is 21.2 m/s.
4	24.8 m/s	Energy gives v = square root of (225 + 392.4) = 24.8 m/s, matching the component method.
5	91.7 m	Range is 900 / 9.81 = 91.7 m, the furthest possible for that speed.
6	They land together	The vertical motion is identical, since a horizontal velocity does not change the vertical acceleration.
7	1.51 m	t = square root of (3.6 / 9.81) = 0.61 s, so x = 2.5 x 0.61 = 1.51 m.
8	277 m	The fall takes 0.55 s, so x = 500 x 0.55 = 277 m, showing that height alone fixes the time.
9	7.82 m, so it hits the wall	h = x tan theta – g x squared over 2 u squared cos squared theta = 51.96 – 44.15 = 7.82 m, below 10 m, so the ball hits the wall.
10	They land together, both accelerating at 9.81 m/s2 downwards	A zero initial velocity does not mean zero acceleration; gravity acts on both balls alike, so their vertical motion and landing time are identical.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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