DP Physics (first assessment 2025)

Question 22N.1A.SL.TZ0.4

Date	November 2022	Marks available	[Maximum mark: 1]	Reference code	22N.1A.SL.TZ0.4
Level	SL	Paper	1A	Time zone	TZ0
Command term		Question number	4	Adapted from	N/A

[Maximum mark: 1]

22N.1A.SL.TZ0.4

A projectile is launched with a velocity $u$ at an angle $θ$ to the horizontal. It reaches a maximum height $s$ . What is the time taken to reach the maximum height?

A. $\frac{2 s}{u \cos θ}$

B. $\frac{2 s}{g}$

C. $\frac{u \cos θ}{g}$

D. $\frac{u \sin θ}{g}$

[1]

Markscheme

Examiners report

This question is very easily answered by elimination. The time to reach max height is the same as for a projectile launched vertically with initial speed $u \sin θ$ , and no other alternative than D contains the term $u \sin θ$ .

Syllabus sections

A. Space, time and motion

A. Space, time and motion » A.1 Kinematics

Question 22N.1A.SL.TZ0.4

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