# Speed of a ball

1. Dec 14, 2009

### schyuler2

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

2. Dec 14, 2009

### PhanthomJay

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

3. Dec 14, 2009

### 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
???

4. Dec 14, 2009

### PhanthomJay

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.

5. Dec 14, 2009

### 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

6. Dec 14, 2009

### PhanthomJay

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 _______?

7. Dec 14, 2009

### 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

8. Dec 14, 2009

### ideasrule

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