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phy 121
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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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?
v0 = 20cm/s
ds = 120cm
a = 20cm/s
????? I am not sure how to get these 3 things for the vertical direction or for the horizontal below
20 cm/s is not an acceleration
The vertical acceleration of a freely falling object is 980 cm/s^2, or 9.8 m/s^2. With the v0 and `ds you correctly identified, you will be able to find vf and `dt for the vertical motion.
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
dt = sqrt(2`ds / a)
= sqrt(2 * 120cm / 20cm/s)
=3.64 s
vf = v0 + a *dt
= 20cm/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?
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
It could be for a short instance of time if it bounces.
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Why does this analysis stop at the instant of impact with the floor?
Because that is when the velocity, acceleration, everything stops at that instant. If it bounces then it would be starting all over again.
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30 min
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You got an incorrect acceleration at the beginning, but otherwise you were mostly on the right track. You should revise.
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Solution
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