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Phy 121
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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 = 20 cm/s, `ds = 120 cm, a = 9.8 cm/s^2 downward
@& 9.8 m/s^2 is 980 cm/s^2. That is the vertical acceleration.
This throws your calculation of vf off by a considerable factor.
Otherwise your reasoning is very good.*@
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf = sqrt(20 cm/s^2 + 2 * 9.8 cm/s^2 * 120 cm) = 52 cm/s,
vAve = (52 cm/s + 20 cm/s) / 2 = 36 cm/s, so `ds =
`dv = 52 cm/s - 20 cm/s = 32 cm/s, so `dt = 32 cm/s / 9.8 cm/s^2 = 3.27 s, so..
`ds = 36 cm/s * 3.27 s = 118 cm (approx.)
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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?
a = -9.8 cm/s^2, v0 = 80 cm/s, `dt = 3.27 s
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• What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
v0 = 0 cm/s
vAve = (80 cm/s + 0 cm/s) / 2 = 40 cm/s
`ds = 40 cm/s * 3.27 s = 131 cm in the horizontal direction
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• After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No. It would depend.
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• Why does this analysis stop at the instant of impact with the floor?
Because we do not know anything about the floor to base any further analysis on.
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