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PHY 231
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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?
Initial velocity is 0 cm/s
Displacement is 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=2*-980 cm/s^2 *120cm
vf=484.97 cm/s
484.97 cm/s is the change in velocity
242.29 cm/s is the average velocity
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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 acceleration is 980 cm/s^2
The ball’s initial velocity is 80 cm/s
180cm/(242.29 cm/s)=0.7423 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?
vf=80 cm/s+-980 cm/s^2 *0.7423s
vf=647.453 cm/s
(647.453 cm/s+80 cm/s)/2=363.727 cm/s=vAve
647.453 cm/s-80 cm/s=567.454 cm/s=change in velocity
'ds=80 cm/s*0.7423s+1/2*980 cm/s^2 *(0.7423s)^2=329.379 cm
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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?
Yes
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Why does this analysis stop at the instant of impact with the floor?
The analysis stops because we cannot find the acceleration after the impact with the knowledge we currently have.
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15 min
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