phy201
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed quesiton 13
A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.
For the interval between the end of the ramp and the floor, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
Vertically
V0= 20cm/s
Vf= 0 after 120 cm
'ds= 120cm to the floor
chnage in velocity vf- v0 = 'dv
0cm/s - 20cm/s = -20cm/s
vave= (20cm/s+ 0cm/s)/2 = 10cm/s
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What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
'dt= 'ds/vave
'dt = 120cm/ 10cm/s
'dt= 12 seconds
V0 =80cm/s
a= 'dv/'dt
a= 80cm/12s
a= 6.7cm/s^2
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
vave= 80cm/s
ds= vave * 'dt
'ds= 80cm/s * 12
'ds= 960 cm
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Yes
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Why does this analysis stop at the instant of impact with the floor?
Because the ball will be at rest
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45 min
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I am a little confused on this question, but I answered the best i could :)
Most of your work was well set up, but you assumed a final vertical velocity of 0. However the velocity goes to 0 only after the acceleration changes. During the uniform-acceleration fall, the ball achieves a speed of hundreds of centimeters per second. This makes a huge difference in the final results (e.g., the time of fall is less than a second, not 12 seconds).
&#Please compare your solutions with the expanded discussion at the link
Solution
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