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Phy 241
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 **
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: ->->->->->->->->->->->-> :
v_0 = 15 m/s
g = -10 m/s
h = 12 m above the ground
After one second the ball's velocity is 5 m/s. At 1.5 seconds the ball reaches its highest point and has traveled 15 meters
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Its average velocity for the 1.5 seconds is 7.5 m/s.
The magnitude of its velocity change while coming to rest is 15 m/s, but it hasn't traveled 15 meters.
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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: ->->->->->->->->->->->-> :
t=0, v_0 =15 m/s, s = 0
t=1, v_1 = 5 m/s, s= 10 m
t= 2, v_2 = -5 m/s, s= 10 m
t= 3, v_3 = -15 m/s, s= -5 m
t=4, v_4 = -25 m/s, s= -20 m
Somewhere between 3 and 4 seconds the ball hits the ground. Assuming uniform downward acceleration, the ball reaches -12 meters at 3.4 seconds at a velocity of 19 m/s in the downward direction
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
when t = 1 and t = 2. The difference is t=2 is in the downward direction
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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: ->->->->->->->->->->->-> :
the ball is 20 meters above the ground when t = .8 seconds
The ball hits the ground at 3.4 s, so in order to find how high the ball is at the 6th second, we would have to figure deflection and other principles as well.
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20 minutes
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4/7/2012 310pm
A child in a car tosses a ball upward so that after release it requires 1/2 second to rise and fall back into the child's hand at the same height from which
it was released. The car is traveling at a constant speed of 10 meters / second in the horizontal direction.
Between release and catch, how far did the ball travel in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
So in 1/2 a second the ball rose and fell. That means the ball traveled 5 meters. Of course to the child, the ball did move horizontally at all.
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As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
answer/question/discussion: ->->->->->->->->->->->-> :
sorta like half of an ellispe concave down
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Sketch the path of the ball as observed by a line of people standing along the side of the road. Describe your sketch. What was shape of the path of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
sorta like half an ellipse concave down
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How fast was the ball moving in the vertical direction at the instant of release? At that instant, what is its velocity as observed by a line of people
standing along the side of the road?
answer/question/discussion: ->->->->->->->->->->->-> :
'dt = .25s
a = -9.81 m/s^2
a = 'dv/'dt
'dv = a*'dt
= -9.81 m/s^2 * .25s
= -2.45 m/s
It would actually be positive because the kid is throwing the ball. so there would be a positve velocity of 2.45 m/s
the vertical velocity by the bystanders would be the same, but since the car is moving, the ball would look like its moving too
So the ball has a horizontal velocity as well of 10 m/s
v = sqrt[(2.45m/s)^2 + (10m/s)^2]
= 10.3 m/s
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How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> :
v_ave = 'ds/'dt
'ds = v_Ave*'dt
= (2.45/2) m/s*.25s
= 0.31 m
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&#Good responses. Let me know if you have questions. &#