Why split a launch into two components?

The horizontal part is uniform and the vertical part accelerates, so solve them separately and combine at the end.

Why does the range formula use sin 2 theta?

Range is u squared sin 2 theta over g because the horizontal speed and the flight time each carry one trigonometric factor.

S4 Physics projectile motion paper 2: angled launches

2026-11-10

Tutor Prep Guide · S4 Physics Projectile Motion

Ten angled launch items: split the velocity into components first, then find time, height and range.

Direct answers　Items 01 to 03 cover time, range and height, 04 to 06 components and the 45 degree range, 07 to 09 energy, complementary angles and landing speed, and 10 the claim that 45 degrees maximises height.

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 is launched at 25 m/s at 30 degrees above the horizontal. Find its time of flight.
FoundationFind the horizontal range of the same 25 m/s launch at 30 degrees.
FoundationHow high does the same ball rise?
IntermediateA ball is launched at 30 m/s at 40 degrees. Find the horizontal component of its velocity.
IntermediateA ball is launched at 20 m/s at 45 degrees. Find its range on level ground.
IntermediateA ball is launched at 30 m/s at 45 degrees. Find its time of flight.
ChallengeA ball is launched at 15 m/s at 60 degrees. Find the maximum height.
ChallengeA ball launched at 25 m/s at 30 degrees lands at the same height. Find its landing speed.
ChallengeWith the same speed, which gives the longer range, 30 degrees or 60 degrees?
Error redoA student claims 45 degrees gives the greatest height. Give the correct conclusion.

03　Answers and one-line notes

#	Answer	One-line note
1	2.55 s	The vertical component is 25 sin 30 = 12.5 m/s, so t = 25 / 9.81 = 2.55 s, with equal rise and fall times.
2	55.2 m	Range is u squared times sin 2 theta over g, giving 55.2 m; the doubled angle is the key.
3	7.96 m	At the top the vertical speed is zero, so H = 12.5 squared over 19.62 = 7.96 m.
4	23.0 m/s	The horizontal part is 30 cos 40 = 23.0 m/s; cosine gives the horizontal part and sine the vertical part.
5	40.8 m	At 45 degrees sin 90 is 1, so the range is 400 / 9.81 = 40.8 m, the largest possible for that speed.
6	4.32 s	The vertical part is 21.2 m/s, so the time is 2 x 21.2 / 9.81 = 4.32 s.
7	8.60 m	The vertical part is 12.99 m/s, so H = 12.99 squared over 19.62 = 8.60 m; a steeper launch rises higher.
8	25 m/s	With no air resistance energy is conserved, so the landing speed equals the launch speed of 25 m/s, now directed 30 degrees below the horizontal.
9	They are equal	sin 60 equals sin 120, so both give the same range; the steeper one simply rises higher and stays longer.
10	45 degrees maximises range only; height grows as the angle nears 90 degrees	Height depends on the square of u sin theta, so it grows all the way to 90 degrees; 45 degrees is special for range alone.

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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