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?

University Physics Volume 1
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Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter3: Motion Along A Straight Line
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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.
Transcribed Image Text: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
Transcribed Image Text: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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