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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, what are the ball's initial velocity, displacement and acceleration in the vertical direction?
v0 = 20cm/s (given in problem)
`ds = 120 cm (given)
a = 980 cm/s^2 (gravity)
What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
`ds = 120 cm (given)
`dv = vf - v0 = 485 - 20 = 465 cm/s
vAve = v0 + vf/2 = 20 + 485 / 2 = 252 cm/s
vf^2 = v0^2 + 2a`ds = (20)^2 + (2*980*120) = `sqrt 235600 = 485 cm/s
What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?
a = 0
v0 = 80 cm/s (given)
'dt = `ds/vAve = 120cm/252cm/s = 0.48s
What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
vf = 80 cm/s
vAve = 80cm/s
`dv = 0 m/s
`ds = vAve`dt = (80 cm/s)(0.48s) = 38 cm
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No.
Why does this analysis stop at the instant of impact with the floor?
Because acceleration is no longer constant, no longer uniform, at least with respect to the conditions of this problem.
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20min.
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Please respond relative to goal of C.
&#This looks very good. Let me know if you have any questions. &#