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12.27 Free Fall and Vertical Motion Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Physics Science worksheet provides structured practice on 12.27 Free Fall and Vertical Motion. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

Free Fall and Vertical Motion: Analyze vertical motion under gravitational acceleration, including launch speed, time in air, maximum height, and impact speed. Aligned to NGSS, CCSS Literacy and Mathematics, AP Physics, and selected state frameworks 12.27 - Free Fall and Vertical Motion: NGSS HS-PS2-1 and SEPs; CCSS HSN-Q.A, HSA-CED.A, HSF-IF, HSF-BF, HSS-ID; AP Physics 1 Unit 1; AP Physics C: Mechanics kinematics enrichment; TX TEKS Physics §112.45(c) motion; FL SC.912.P.12.2; VA PH.2; MA/CA/NY HS-PS2.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Physics

Subject

Science

Domain

Science Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related 12.27 Free Fall and Vertical Motion skill assessments:

Problem 1Difficulty: Mastery
A platform is high enough that a ball launched upward at 8.0 m/s hits the ground with twice its launch speed. How high is the platform? Use g = 9.8 m/s2.
Aabout 3.3 m
Babout 6.5 m
Cabout 13.1 m
Dabout 9.8 m
Solution & Explanation:Set v_impact = 2v0. Then (2v0)2 = v02 + 2gh, so h = 3v02/(2g) = 3(64)/(19.6) = about 9.8 m.
Problem 2Difficulty: Mastery
An object launched upward from ground level has vertical speed 15.0 m/s when it is 12.0 m above the launch point. What maximum height will it reach? Use g = 9.8 m/s2.
Aabout 15.0 m
Babout 18.8 m
Cabout 23.5 m
Dabout 27.3 m
Solution & Explanation:From the 12.0 m point, the remaining rise is v2/(2g) = 15.02/19.6 = about 11.5 m. Add 12.0 m to get about 23.5 m.
Problem 3Difficulty: Mastery
Use the quadratic fit in the image: y = 1.2 + 9.6t - 4.8t2. What gravitational acceleration magnitude does this data model imply?
A4.8 m/s2
B9.6 m/s2
C8.4 m/s2
D19.2 m/s2
Solution & Explanation:In vertical motion, the coefficient of t2 is -1/2 g. Since -1/2 g = -4.8, the model implies g = 9.6 m/s2.

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Frequently Asked Questions

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QIs this 12.27 Free Fall and Vertical Motion worksheet free for teachers and homeschoolers?

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Physics Science students. It aligns with educational standard 12.27 (NGSS HS-PS2-1 and SEPs; CCSS HSN-Q.A, HSA-CED.A, HSF-IF, HSF-BF, HSS-ID; AP Physics 1 Unit 1; AP Physics C: Mechanics kinematics enrichment; TX TEKS Physics §112.45(c) motion; FL SC.912.P.12.2; VA PH.2; MA/CA/NY HS-PS2.).

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