Activity 1: Problem SolvingDirections: Show your complete solutions for the following worded problems.1. A flower pot falls a window of a three-storey apartment. How far has it fallen after 2seconds? What is the pot's velocity after I s of fall? after 2 seconds of fall?2. A brick falls freely from a high scaffold and hits the ground after 3.5 seconds. Howhigh is the scaffold? What is the velocity of the brick just before hitting the ground?3. Alice thrown a ball upwards vertically from the roof of a building. The ball leaves herhand at a point level with the roof railing with an upward speed of 15 m/s. On its wayback down, it just misses the railing. Find (a) the ball's position and velocity 1.0 s and4.0 s after leaving her hand; (b) the ball's velocity when it is 5.0 m above the railing;(c) the maximum height reached. olem 1.A stone is dropped from the top of a building and falls freely from rest. What isits position (height) and its velocity after 1.0s, 2.0s, and 3.0s?To find for the position (height), we willuse:Know that initial velocity (vi) = 0a. Find h after 1.0sh=0(1.0s)-(9.8 m/s)(1.0s)22|h=v₁t - 9²2h=0-9.8m2h=-4.9 m(negative sign denotes downward direction)b. Find h after 2.0sh=0(2.0s) - (9.8 m/s²)(2.0s)²2h=0-39.2 m2h = -19.6 m or 19.6 m downwardGiven: v₁ = 20 m/sFind:a. v after 1sb. v after 5sc. d after 1sd. t from origin to maximum heightd from origin to maximum heighte.Solutions:a.y after isuse:V₁ = v₁ - gtvy=20 m/s-(9.8 m/s²)(1.0s)vy= 20 m/s -9.8 m/svy=10.2 m/sb. v after 2suse:Sample Problem 2.A boy threw a ball upward with an initial velocity of 20 m/s and was able to catch itbefore it reached the ground on its return.a. What was the ball's velocity after 1s?b. What was the ball's velocity after 2s?c. What was its displacement after 1s?d. How long did it take the ball to reach its maximum height?e. How far was the maximum height from the starting point?V₁ = V₁ - gtvy-20 m/s-(9.8 m/s²)(2.0s)v=20 m/s-19.6 m/svy=0.4 m/sc. d after 1suse:c. Find h after 3.0sTo find for velocity (vr), we will use:Vf=v₁-gth=0(3.0s)-(9.8 m/s²)(3.0s)²d=v₁t - 92²d=20m/s(1.0s)-(9.8 m/s2)(1.0s)²2d=20m-4.9 md=15.1 m(positive sign denotes that the object ismoving upward)h=0-88.2 m2h=-44.1 m or 44.1 m downwarda. Find vrafter 1.0svy=0-(9.8 m/s²)(1.0s)v=-9.8 m/sb. Find vy after 2.0s2vy=0-(9.8 m/s²)(2.0s)v=-19.6 m/sc. Find vy after 3.0svy=0-(9.8 m/s²)(.Os)vy=-29.4 m/sd. t from origin to maximum heightNote: An object thrown upward willreach a certain point (maximumheight) where it will momentarilystop and then start todownward. At this point, vy= 0.moveV₁ = V₁-gtManipulating the equation,V₂ - Vf9Use:t=t=20 m/s-09.8 m/s²t=2.04 se. d from origin to maximum heightuse:gt²d=v₁t-d=20m/s (2.04 s)-(9.8 m/s2)(2.04s)22d = 40.8 m-20.40 md=20.4 m

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Description: Activity 1: Problem SolvingDirections: Show your complete solutions for the following worded problems.1. A flower pot falls a window of a three-storey apartment. How far has it fallen after 2seconds? What is the pot's velocity after I s of fall? aft

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1. A flower pot falls a window of a three-storey apartment. How far has it fallen after 2 seconds? What is the pot's velocity after I s of fall? after 2 seconds of fall?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Displacement, Velocity and Acceleration

In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.

Linear Displacement

The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.

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Question

Activity 1: Problem Solving
Directions: Show your complete solutions for the following worded problems.
1. A flower pot falls a window of a three-storey apartment. How far has it fallen after 2
seconds? What is the pot's velocity after I s of fall? after 2 seconds of fall?
2. A brick falls freely from a high scaffold and hits the ground after 3.5 seconds. How
high is the scaffold? What is the velocity of the brick just before hitting the ground?
3. Alice thrown a ball upwards vertically from the roof of a building. The ball leaves her
hand at a point level with the roof railing with an upward speed of 15 m/s. On its way
back down, it just misses the railing. Find (a) the ball's position and velocity 1.0 s and
4.0 s after leaving her hand; (b) the ball's velocity when it is 5.0 m above the railing;
(c) the maximum height reached.

olem 1.
A stone is dropped from the top of a building and falls freely from rest. What is
its position (height) and its velocity after 1.0s, 2.0s, and 3.0s?
To find for the position (height), we will
use:
Know that initial velocity (vi) = 0
a. Find h after 1.0s
h=0(1.0s)-(9.8 m/s)(1.0s)2
2
|h=v₁t - 9²
2
h=0-9.8m
2
h=-4.9 m
(negative sign denotes downward direction)
b. Find h after 2.0s
h=0(2.0s) - (9.8 m/s²)(2.0s)²
2
h=0-39.2 m
2
h = -19.6 m or 19.6 m downward
Given: v₁ = 20 m/s
Find:
a. v after 1s
b. v after 5s
c. d after 1s
d. t from origin to maximum height
d from origin to maximum height
e.
Solutions:
a.
y after is
use:
V₁ = v₁ - gt
vy=20 m/s-(9.8 m/s²)(1.0s)
vy= 20 m/s -9.8 m/s
vy=10.2 m/s
b. v after 2s
use:
Sample Problem 2.
A boy threw a ball upward with an initial velocity of 20 m/s and was able to catch it
before it reached the ground on its return.
a. What was the ball's velocity after 1s?
b. What was the ball's velocity after 2s?
c. What was its displacement after 1s?
d. How long did it take the ball to reach its maximum height?
e. How far was the maximum height from the starting point?
V₁ = V₁ - gt
vy-20 m/s-(9.8 m/s²)(2.0s)
v=20 m/s-19.6 m/s
vy=0.4 m/s
c. d after 1s
use:
c. Find h after 3.0s
To find for velocity (vr), we will use:
Vf=v₁-gt
h=0(3.0s)-(9.8 m/s²)(3.0s)²
d=v₁t - 92²
d=20m/s(1.0s)-(9.8 m/s2)(1.0s)²
2
d=20m-4.9 m
d=15.1 m
(positive sign denotes that the object is
moving upward)
h=0-88.2 m
2
h=-44.1 m or 44.1 m downward
a. Find vrafter 1.0s
vy=0-(9.8 m/s²)(1.0s)
v=-9.8 m/s
b. Find vy after 2.0s
2
vy=0-(9.8 m/s²)(2.0s)
v=-19.6 m/s
c. Find vy after 3.0s
vy=0-(9.8 m/s²)(.Os)
vy=-29.4 m/s
d. t from origin to maximum height
Note: An object thrown upward will
reach a certain point (maximum
height) where it will momentarily
stop and then start to
downward. At this point, vy= 0.
move
V₁ = V₁-gt
Manipulating the equation,
V₂ - Vf
9
Use:
t=
t=20 m/s-0
9.8 m/s²
t=2.04 s
e. d from origin to maximum height
use:
gt²
d=v₁t-
d=20m/s (2.04 s)-(9.8 m/s2)(2.04s)2
2
d = 40.8 m-20.40 m
d=20.4 m

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