cq_1_081

#$&*

PHY 121

Your 'cq_1_08.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** CQ_1_08.1_labelMessages **

A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the

acceleration of gravity is 10 m/s^2 downward.

What will be the velocity of the ball after one second?

answer/question/discussion: ->->->->->->->->->->->-> :

After one second, the ball should have lost 10 m/s, so it should be traveling 15 m/s.

#$&*

What will be its velocity at the end of two seconds?

answer/question/discussion: ->->->->->->->->->->->-> :

At the end of 2 seconds, it should have lost another 10 m/s, so it should be traveling 5

m/s.

#$&*

During the first two seconds, what therefore is its average velocity?

answer/question/discussion: ->->->->->->->->->->->-> :

The initial velocity is 25 m/s and the final velocity after 2 seconds is 5 m/s. To find

the average, you add the two numbers and divide by 2. 25 m/s + 5 m/s = 30 m/s. 30 m/s

/ 2 = 15 m/s

The average velocity is 15 m/s for the first 2 seconds.

#$&*

How far does it therefore rise in the first two seconds?

answer/question/discussion: ->->->->->->->->->->->-> :

It rises 30 m in the first two seconds.

#$&*

What will be its velocity at the end of a additional second, and at the end of one more

additional second?

answer/question/discussion: ->->->->->->->->->->->-> :

After an additional second, it will have reached its peak and will be starting back

down. This gives it a velocity of 5 m/s - 10 m/s = -5 m/s.

At the end of one more second, it will be going -5 m/s - 10 m/s = -15 m/s.

#$&*

At what instant does the ball reach its maximum height, and how high has it risen by

that instant?

answer/question/discussion: ->->->->->->->->->->->-> :

It will reach its maximum height at 2.5 seconds. This is because it changes direction

halfway between 2 and 3 seconds. It will be at 13.25 m at that time. To get that

answer, I averaged the velocity starting at 25 m/s and ending at 0 m/s. This gave me an

average velocity of 12.5 m/s. At 12.5 m/s times 2.5 s, the height is 31.25 m.

However,the picture that I drew, shows the following:

Second 1 - v0 = 25 m/s, vf = 15 m/s, vAve = 20 m/s, for a total of 20 m in that second.

Second 2 - v0 = 15 m/s, vf = 5 m/s, vAve = 15 m/s, for a total of 15 m in that second

Second 3 - v0 = 5 m/s, vf = -5 m/s, vAve doesn't help, but it's going up 5 meters and

then back down 5 meters at this time. This total is 45 meters.

#$&*

What is its average velocity for the first four seconds, and how high is it at the end

of the fourth second?

answer/question/discussion: ->->->->->->->->->->->-> :

The average velocity for the first four seconds is 5 m/s. You take the 25 m/s and add

the -15 m/s, to get 10 m/s and divide that by 2.

#$&*

How high will it be at the end of the sixth second?

answer/question/discussion: ->->->->->->->->->->->-> :

I think it's done by the end of the sixth second. My sketch shows it's done at 5

seconds.

#$&*

*#&!*#&!

&#This looks very good. Let me know if you have any questions. &#