cq_1_082

#$&*

phy 201

Your 'cq_1_08.2' 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_08.2_labelMessages **

Seed 8 2

A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).

• How high does it rise and how long does it take to get to its highest point?

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

t0=12 m

a= -10m/s/s

v0=15 m/s

vf=0 m/s

• vf = v0 + a * `dt

vf-v0)/a=dt

-15/-10=dt

Dt=1.5s

vAve=7.5 m/s *1.5 s

it rises 11.25 m to a maximum height of 26.25 meters and it takes 1.5 seconds to get there

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Good. It does rise 11.25 m. But having started from the 12 m position it rises only to a 23.25 m height.

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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?

Ds=-26.25 m

ds = (v0 + vf) / 2 * `dt

vf = v0 + a * `dt

vf= 15 m/s -10m/s/s *4 s

vf= -25 m/s

ds-vAve=dt

?????

I don’t know how to do this

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

#$&*

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Good try.

The ball started from the 12 m position so its displacement, from that position, is only -12 m. You also don't know that it takes 4 seconds to get there.

What you do know, therefore, is that

`ds = -12 m

v0 = +15 m/s

a = -10 m/s^2.

Now which equation(s) contain these three variables, and which will be the easiest to solve for the remaining variable?

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• At what clock time(s) will the speed of the ball be 5 meters / second?

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

5m/s = 15 m/s-10 m/s^2 *dt

5m/s-15m/s)/-10=dt

At 1 second the speed will be at 5m/s

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That's true. So far so good.

But there's another time at which the speed is 5 m/s. Hint: When the velocity is -5 m/s, the speed is 5 m/s.

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• At what clock time(s) will the ball be 20 meters above the ground?

ds = (v0 + vf) / 2 * `dt

8 - 7.5)*2= 1 second

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

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

• `ds = v0 `dt + .5 a `dt^2

Ds= 15 m/s (6) +.5(-10m/s^2)(6^2)

90 + -180

-90 m is how high it would be. Im assuming it would’ve already come down and it the ground,

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*#&!

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Good attempts. As you know you didn't get everything, especially on the first problem. I don't think you'll have much trouble going a little further.

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