cq_1_131

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PHY201

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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?

Horizontal is never changing at 80 cm/s

v0=20 cm/s vertical

9.8m/s = 980cm/s

0 seconds = 20 cm/s

1 second = 20+980=1000 cm/s

2 seconds = 1000+980= 1980cm/s

120=v0'dt+.5a'dt^2

120=20(2)+.5(a)2^2

120=40+2a

80=2a

a=40cm/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=20^2 + 2(40)(120)

vf^2= 400+9600

vf^2=10000

vf=100 cm/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?

Horizontal has no acceleration as it stays constance. 80cm/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?

120=80('dt)

1.5='dt

vf^2=80^2

vf=80

vAve=80cm/s

nochange in velocity

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After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?

yes in the opposity direction

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Why does this analysis stop at the instant of impact with the floor?

Gravity has a constant 9.8 m/s on the ball but when it hits the floor we do not know that the ball will do. espcially not know what it is hitting what kind of ball it is.

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30 minutes

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