Speed of a ball

1. The problem statement, all variables and given/known data
A ball is thrown straight upward from the ground with a speed of 30 m/s. What is the average speed of the ball in the first 5 seconds?


2. Relevant equations
VBy2 = VAy2 + 2(ay) * dABy

VBy = VAy + ay * change in time

Average velocity = dABy / change in time

3. The attempt at a solution
VBy2 = VAy2 + 2(ay) * dABy
(0)2 = (30)2 + 2(9.8) * dABy
0 = 900 + 19.6 * daby
-900 = 19.6 *dABy
-45.9 = dABy
^but dABy can't be negative. what am i doing wrong?

also-- do i use "5 sec" for my time interval on the other equations? i cannot figure out what it means by "first 5 seconds." HELP
 

PhanthomJay

Science Advisor
Homework Helper
Gold Member
6,996
401
1. The problem statement, all variables and given/known data
A ball is thrown straight upward from the ground with a speed of 30 m/s. What is the average speed of the ball in the first 5 seconds?


2. Relevant equations
VBy2 = VAy2 + (that should be a minus sign, not a plus sign)2(ay) * dABy

VBy = VAy + ay * change in timewatch your minus signs

Average velocity = dABy / change in time
the problem asks for average speed, which is not the same as average velocity
3. The attempt at a solution
VBy2 = VAy2 + 2(ay) * dABy
(0)2 = (30)2 + 2(9.8) * dABy
0 = 900 + 19.6 * daby
-900 = 19.6 *dABy
-45.9 = dABy
^but dABy can't be negative. what am i doing wrong?
once you make your minus sign correction, 45.9 m is the distance traveled up. How long did it take to reach that distance? Where will the ball be after 5 seconds? What total distance will the ball have travelled in that 5 seconds (45.9 meters up and an additional distance down)? Then average speed is ???
 
well if it travels a total of 2*45.9m ... then the dABy = 91.8

so:
average velocity = dABy / t
= 91.8/5
=18.36 m/sec
???
 

PhanthomJay

Science Advisor
Homework Helper
Gold Member
6,996
401
well if it travels a total of 2*45.9m ... then the dABy = 91.8

so:
average velocity = dABy / t
= 91.8/5
=18.36 m/sec
???
No, first off, you are looking for average speed, after 5 seconds. You can use y = V_yo(t) - 1/2gt^2 (where t=5) to find out the position of the ball after 5 seconds ( it'll be on its descent at that point). Then calculate the average speed by first calculating the total distance it has traveled during that time.
 
Thanks! so,
y =V_yo(t) - 0.5gt^2
y = 30(5) - 0.5 (9.8) (5)^2
y = 150 - 4.9 * 25
y = 150 - 122.5
y = 27.5 m

then...
Vavg = dABy / t
Vavg = 27.5 m /5 sec
Vavg = 5.5 m/sec
 

PhanthomJay

Science Advisor
Homework Helper
Gold Member
6,996
401
Thanks! so,
y =V_yo(t) - 0.5gt^2
y = 30(5) - 0.5 (9.8) (5)^2
y = 150 - 4.9 * 25
y = 150 - 122.5
y = 27.5 m

then...
Vavg = dABy / t
Vavg = 27.5 m /5 sec
Vavg = 5.5 m/sec
You are getting close, but again, the problem is asking for average speed, not average velocity. What you have calculated is the average velocity, which is displacement divided by time, a vector quantity. Average speed is distance traveled divided by time, a scalar quantity. It travels 45.9 m up and then (45.9 - 27.5) m down, for a total distance of ____? and an average speed of _______?
 
Total distance: 45.9 + 23.4 = 63.9 m

avg speed = dist/time
avg speed = 63.9 m / 5 sec
avg speed = 13.9 m/sec^2
 

ideasrule

Homework Helper
2,263
0
45.9-27.5 isn't 23.4. Also, why's your average speed in m/s^2 instead of m/s?
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top