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PHY 121
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.
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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: ->->->->->->->->->->->-> :
To get time, we can use the equation, t = v0/a. So 15m/s / 10m/s^2 = 1.5s. . Then you can take that and use this formula:
h = h0 + v0(t)+ ½ *a(t)^2 , so I can substitute everything I have in. h0 is the distance from earth, so that would be 0. v0 is 15m/s. A is -10m/s^2. T is 1.5s. So h = (0+15(1.5) + ½(-10)(1.5)^2 = 11.25 m. Then you add 12 meters to the height because we started 12 meters above ground to get 23.25m.
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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?
answer/question/discussion: ->->->->->->->->->->->-> :
It should be going -21.3m/s when it hits the ground.I used the equation vf^2 = v0^2 + 2(a)(`ds) So I fill in the numbers as vf = sqrt(2(-9.8)(23.25) = -21.3m/s. For the time, you can use the equation 1/2(a)(t)^2 + v0(t) This is the equation from the max height of the ball, so after solving this, you will add 1.5 seconds to it. Which equals 3.68 seconds
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
To find the clock time at 5 m/s . You can use the equation t = `dv/a. So `dv is -10m/s / -9.8 which equals 1.02 s.
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At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
You can find out the time at 20 meters by using the equation, t = sqrt((2(8m))/-9.8 which is 1.26 seconds.
To find the height of the ball after 6 seconds, you can use the equation, `ds = 1/2(v0+v0+a(t)) t. So if you fill in the variables, you get `ds = 1/2(15+15+-9.8(6)) * 6 which equals -86.4 m.
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@& Very good job with the equaitons.
Also check the discussion. No revision is necessary.*@
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Solution
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