Calculating Average Speed of Ball in First 5 Seconds

[schyuler2]

Homework Statement

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?

Homework Equations

VBy² = VAy² + 2(ay) * dABy

VBy = VAy + ay * change in time

Average velocity = dABy / change in time

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]

Homework Statement

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?

Homework Equations

VBy² = VAy² + (that should be a minus sign, not a plus sign)[/color]2(ay) * dABy

VBy = VAy + ay * change in timewatch your minus signs[/color]

Average velocity = dABy / change in time

the problem asks for average speed, which is not the same as average velocity

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 traveled in that 5 seconds (45.9 meters up and an additional distance down)? Then average speed is ?

 

[schyuler2]

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]

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.

 

[schyuler2]

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]

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 _______?

 

[schyuler2]

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]

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