Why can the horizontal and vertical motions be treated separately?

Only time links the two directions: horizontally the speed is constant and vertically the acceleration is constant, so work them out separately and then combine.

Why is the falling time of a horizontal throw independent of its speed?

The vertical initial speed is zero, so the falling time depends only on the height while the horizontal speed only changes the range.

Why check units again in a drill paper?

Projectile slips often come from centimetres left as centimetres or grams left as grams, so fix the units before substituting.

Is 45 degrees always the best launch angle?

Without air resistance and on level ground 45 degrees gives the maximum range, but when launching to a lower level the best angle is under 45 degrees.

S4 Physics projectile motion paper 4: ten common slips

2026-11-10

Tutor Prep Guide · S4 Physics Projectile Motion

Ten items rewritten from the usual slips: mixing components, dropping the factor of two and treating a horizontal speed as a vertical one.

Direct answers　Items 01 to 03 retry time, range and vertical speed for horizontal throws, 04 to 06 reverse work plus angled time and range, 07 to 09 peak height, a wall and a table edge, and 10 dividing a horizontal speed by g.

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 stone is thrown horizontally at 12 m/s from a height of 5.0 m. With g = 9.81 m/s², find the time taken to reach the ground.
FoundationUsing the same throw, find the horizontal distance travelled by the stone.
FoundationA ball is thrown horizontally at 20 m/s. What is its vertical speed 2.0 s later?
IntermediateA ball thrown horizontally lands 24 m from the release point, which is 20 m above the ground. Find the horizontal speed.
IntermediateA ball is launched at 25 m/s at 35 degrees above the horizontal. Find its time of flight back to the same height.
IntermediateFor the same launch, find the horizontal range.
ChallengeA ball is launched at 30 m/s at 50 degrees above the horizontal. Find the maximum height above the launch point.
ChallengeA ball is launched at 15 m/s at 45 degrees towards a 5 m high wall 12 m away. Find its height when it reaches the wall and say whether it clears it.
ChallengeA ball rolls horizontally off a table 1.2 m high at 4.0 m/s. Find the falling time and the horizontal distance.
Error redoA ball is thrown horizontally at 15 m/s from 20 m above the ground. A student answers 15 / 9.81 = 1.53 s. What is wrong, and what is the correct time?

03　Answers and one-line notes

#	Answer	One-line note
1	1.01 s	The vertical initial speed is zero, so t = sqrt(2h/g) = sqrt(2 x 5.0 / 9.81) = 1.01 s; the horizontal speed does not affect the falling time.
2	12.1 m	The horizontal motion is uniform, so x = u t = 12 x 1.01 = 12.1 m; quoting 5.0 m as the range is the usual slip.
3	19.6 m/s downwards	v = g t = 9.81 x 2.0 = 19.6 m/s, which is independent of the 20 m/s horizontal throw.
4	11.9 m/s	Find the falling time first, t = sqrt(2 x 20 / 9.81) = 2.02 s, then u = 24 / 2.02 = 11.9 m/s; doing it the other way round leads to the wrong answer.
5	2.92 s	T = 2u sin(theta) / g = 2 x 25 x sin35 / 9.81 = 2.92 s; dropping the factor of two gives 1.46 s, which is only the time to the peak.
6	59.9 m	R = u cos(theta) x T = 25 x cos35 x 2.92 = 59.9 m, using the horizontal component rather than the full 25 m/s.
7	26.9 m	H = (u sin(theta))^2 / (2g) = (30 x sin50)^2 / 19.62 = 26.9 m; using the full 30 m/s overestimates the height.
8	5.72 m, so it clears the wall	t = 12 / (15 cos45) = 1.13 s, then y = 10.6 x 1.13 – 4.905 x 1.13^2 = 5.72 m, which is above the 5 m wall.
9	0.49 s, 1.98 m	t = sqrt(2 x 1.2 / 9.81) = 0.49 s and x = 4.0 x 0.49 = 1.98 m; the table height is not a vertical initial speed.
10	The horizontal speed was used; the correct time is 2.02 s	The 15 m/s is horizontal and the vertical initial speed is zero, so t = sqrt(2 x 20 / 9.81) = 2.02 s; dividing a horizontal speed by g means nothing.

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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