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PHY 201
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** CQ_1_13.1_labelMessages **
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?
'ds = 0 cm
vertical acceleration is 980 cm/s^2
vertical initial velocity = 0 cm/s
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What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
'ds = 120 cm, a = 980 cm/s^2
vf^2 = v0^2 + 2a'ds, vf = sqrt(2a'ds) = sqrt(2*980 cm/s^2 * 120 cm) = 485 cm/s
vAve = (v0 + vf) / 2 = 485 cm/s / 2 = 243 cm/s
vAve = 'ds / 'dt => 'dt = 'ds / vAve = 120 cm / 243 cm/s = 0.5 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?
a = 0 m/s^2
v0 = 80 cm/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?
final velocity and average velocity = 80 cm/s
'dv = 80 cm/s - 80 cm/s = 0 cm/s #$&*
After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
No #$&*
Why does this analysis stop at the instant of impact with the floor?
Because of other forces? #$&*
&#Very good responses. Let me know if you have questions. &#