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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, hat are the ball's initial velocity, displacement and acceleration in the vertical direction?
The initial velocity is 20 cm/s. The displacement will be 120 cm. The acceleration is 980 cm/s^2.
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf^2 = v0^2 + 2a`ds
vf^2 = (20 cm/s)^2 + 2(980 cm/s)(120 cm)
vf^2 = 400 cm^2/s^2 + 235,200 cm^2/s^2
vf^2 = 235,600 cm^2/s^2
vf = +- 485 cm/s
vf = 485 cm/s
Displacement given as 120 cm.
Change in velocity is 465 cm/s (485 cm/s - 20 cm/s)
Average velocity is 252.5 cm/s (20 cm/s + 485 cm/s)/2
`ds/vAve = `dt
120 cm/252.5 cm/s = .48 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?
The ball's initial velocity in the horizontal direction is given as 80 cm/s.
Horizontally, the initial velocity is 80 cm/s. The acceleration is 980 cm/s^2. It takes .48 seconds for the ball to land.
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Acceleration is in the vertical direction. Gravity doesn't exert a force in the horizontal direction, so there is no acceleration in that direction associated with gravity.
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Acceleration is in the vertical direction. Gravity doesn't exert a force in the horizontal direction, so there is no acceleration in that direction associated with gravity.
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What therefore are its displacement, final velocity, average velocity and change in velocity in the horizontal direction during this interval?
`ds = (v0 + vf)/2 * `dt
vf = v0 + a`dt
vf = 80 cm/s + (980 cm/s*.48 s)
vf = 550.4 cm/s
vAve = (80 cm/s + 550.4 cm/s)/2 = 315.2 cm/s
`ds = 315.2 cm/s * .48 s
`ds = 151.3 cm
Change in velocity is 550.4 cm/s - 80 cm/s = 470.4 cm/s
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
I think that the ground will push back and the ball will accelerate away from the floor. It will do it smoothly, so I would call that uniform acceleration. However, the gravity will continue to pull back and it will eventually (probably) come to rest.
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Acceleration changes drastically on contact with the floor.
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Why does this analysis stop at the instant of impact with the floor?
I am sure there is a way, but I don't know how to determine the trajectory and speed of the ball when it bounces off the floor.
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See my note. You should revise and resubmit this, and it will be a good question to deal with when you return.
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