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Vertical Motion Challenge

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Vertical Motion Challenge

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Quiz on vertical motion problems

Created by

Aimee
Aimee
Philippines

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Vertical Motion Challenge
 

Vertical Motion ChallengeOnline version

Quiz on vertical motion problems

by Aimee
1

A ball is dropped from rest from a height of 19.6 m. How long to reach the ground?

2

A stone is dropped from a bridge and reaches the water after 3.0 s. What is the height of the bridge?

3

A ball is thrown vertically upward at 19.6 m/s. What is its velocity at the highest point?

4

A ball is thrown upward at 19.6 m/s. What maximum height will it reach?

5

A ball is thrown upward at 24.5 m/s. What will its velocity be after 1.5 s?

6

A ball is thrown upward at 19.6 m/s. At what two times will it be 14.7 m above its starting point?

7

A stone is thrown downward at 5.0 m/s from a height of 30 m. Approximately how long to reach the ground?

8

A ball is thrown upward from the top of a 20 m building with 14.7 m/s. What is the maximum height above ground?

9

A ball is thrown upward. After 2.0 s, its velocity is 4.6 m/s downward. What was its initial velocity?

10

A stone is thrown vertically upward and reaches a maximum height of 45 m. What was its initial velocity?

Feedback

From vertical motion equations with g=9.8 m/s2. Source: Directions.6a82947110109.docx.

Uses s=0.5 g t^2 with g=9.8 m/s2. Source: Directions.6a82947110109.docx.

Velocity at top is zero. Source: Directions.6a82947110109.docx.

Use v^2 = u^2 - 2 g h with v=0. Source: Directions.6a82947110109.docx.

v = u - g t ; g=9.8 m/s2. Source: Directions.6a82947110109.docx.

Solve s(t)=14.7 with s= u t - 0.5 g t^2. Source: Directions.6a82947110109.docx.

Use s ≈ ut + 0.5 g t^2 with downward as positive. Source: Directions.6a82947110109.docx.

Height gain = u^2/(2g) ≈ (14.7^2)/(19.6) ≈ 11.0 m. Total ~31.0 m. Source: Directions.6a82947110109.docx.

Using v = u - g t with downward positive. Source: Directions.6a82947110109.docx.

From v^2 = u^2 - 2 g h with h=45 m. Source: Directions.6a82947110109.docx.

 
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