Why is the vertical initial velocity zero in a horizontal throw?

The launch velocity is purely horizontal, so the vertical component is zero and the vertical motion is free fall from rest.

Does the horizontal speed change the time of flight?

No; the two directions are independent and only the height and gravity set the time.

S4 Physics projectile motion paper 1: horizontal throws

2026-11-10

Tutor Prep Guide · S4 Physics Projectile Motion

Ten items on horizontal throws: find the vertical falling time first, then multiply by the horizontal speed.

Direct answers　Items 01 to 03 cover time and range, 04 to 06 vertical speed and reverse work, 07 to 09 components and heights, and 10 treating a horizontal speed as a vertical one.

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 thrown horizontally at 10 m/s from a 20 m tower. Find its time of flight with g = 9.81 m/s2.
FoundationHow far horizontally does the same ball travel before landing?
FoundationA ball is thrown horizontally from 45 m. Find the time of flight.
IntermediateAn object is thrown horizontally. Find its vertical speed after 2 s.
IntermediateA ball leaves a 5 m table horizontally at 8 m/s. Find the horizontal distance to the floor.
IntermediateAn object lands 6 m from the launch point 1.5 s after being thrown horizontally. Find its horizontal speed.
ChallengeAn object is thrown horizontally at 15 m/s. Find its horizontal and vertical speeds and the resultant speed after 3 s.
ChallengeAn object is thrown horizontally at 12 m/s from 80 m. Find the time of flight and the horizontal range.
ChallengeAn object is thrown horizontally from 25 m. Find its vertical speed on landing.
Error redoA ball is thrown horizontally at 10 m/s from 20 m. A student divides 10 by 9.81 and gets 1.02 s; find the correct time.

03　Answers and one-line notes

#	Answer	One-line note
1	2.02 s	Vertical motion starts from rest, so t = square root of (2 x 20 / 9.81) = 2.02 s; the horizontal speed does not change it.
2	20.2 m	The horizontal speed stays constant, so x = 10 x 2.02 = 20.2 m.
3	3.03 s	t = square root of (90 / 9.81) = 3.03 s; quadrupling the height only doubles the time because time goes as a square root.
4	19.6 m/s downwards	Vertical motion accelerates freely, so v = 9.81 x 2 = 19.6 m/s, independent of the horizontal speed.
5	8.08 m	First t = square root of (10 / 9.81) = 1.01 s, then x = 8 x 1.01 = 8.08 m.
6	4 m/s	Horizontal motion is uniform, so u = 6 / 1.5 = 4 m/s.
7	15 m/s, 29.4 m/s and 33.0 m/s	The horizontal component stays 15 m/s, the vertical is 9.81 x 3 = 29.4 m/s, and the resultant is 33.0 m/s.
8	4.04 s and 48.5 m	t = square root of (160 / 9.81) = 4.04 s and x = 12 x 4.04 = 48.5 m.
9	22.1 m/s	v squared is 2 g h, so v = square root of (2 x 9.81 x 25) = 22.1 m/s; it does not depend on the horizontal speed.
10	2.02 s	Only the height sets the time: t = square root of (40 / 9.81) = 2.02 s, and treating the horizontal speed as a vertical one is the classic slip.

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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