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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.
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Seed Questions 1_08.2 submitted 12 Feb 11 around 9:15 AM.
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45 mins
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Seed Questions 1_08.2
You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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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: ->->->->->->->->->->->-> :
Given: v0 = 15 m/s, vf = 0, h = 12 m, (a or g) = -10 m/s^2
dt = (vf - v0) / a
dt = (0 - 15 m/s) / -10 m/s^2
dt = -15 m/s / -10 m/s^2
dt = 1.5 s
ds = (vf + v0) / 2 * dt
ds = (0 m/s + 15 m/s) / 2 * 1.5 s
ds = 15 m/s / 2 * 1.5 s
ds = 7.5 m/s * 1.5 s
ds = 11.25 m
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@& Good. The solution to this question can be obtained by reasoning, but the equations are probably preferable in this case. Having obtained the solution, though, it's always good to spend a minute thinking about how it could have been obtained by direct reasoning.
Some situations cannot be reasoned out directly, and of course you have to be good with the equations to get those.*@
• 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: ->->->->->->->->->->->-> :
Given: v0 = 15 m/s, vf = 0, h = 12 m,( a or g) = -10 m/s^2, dt = 1.5 s, ds = 11.25 m
dv = g * dt
dv = -10 m/s^2 * 1.5 s
dv = -15 m/s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
dt = (vf - v0) / a
dt = (0 - 5 m/s) / -10 m/s^2
dt = -5 m/s / -10 m/s^2
dt = 0.5 s
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@& The initial velocity of the ball for this interval is 15 m/s, the final velocity is 5 m/s.*@
• 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: ->->->->->->->->->->->-> :
ds = (vf + v0) / 2 *dt
20 m = (0 + 15 m/s) / 2 * dt
20 m = 7.5 m/s * dt
dt = 20 m / 7.5 m/s
dt = 2.67 s
@& The ball doesn't start out from the ground, so when it reaches the 20 m position it will not have traveled 20 m.
The ball comes to rest after 1.5 seconds, as can be reasoned directly. So it doesn't have 2.67 s to rise.*@
ds = v0 dt - ½ g dt^2
ds = 15 m/s * 6 s - ½ (10 m/s^2) (6 s)^2
ds = (15 m/s) (6 s) - ½ (10 m/s^2) (6 s)^2
ds = 90 m - ½ (360 m)
ds = 150 m - 180 m
ds = -30 m
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*#&!*#&!
@& A couple of your steps will need revision. Don't spend a lot of time on this; submit what you can do in 10-15 minutes. I expect you'll get it in that time, but if you don't I'll send you a link to a soluion.*@