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Phy 121
Your 'cq_1_13.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** 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?
v0=20cm/s
ds=120cm
a=980cm/s^2
vf^2=v0^2+2*a*ds
vf^2=(20m/s)^2+2*(980cm/s^2)*120cm
vf=sqrt((20m/s)^2+2*(980cm/s^2)*120cm)
vf=235600
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vf=sqrt((20m/s)^2+2*(980cm/s^2)*120cm) = sqrt( 400 m^2 / s^2 + 235 200 m^2 / s^2 ) = sqrt( 235 600 m^2 / s^2) = 480 m/s, very approximately.
Be sure you understand how the units work out.
I know you'll see that you missed taking the square root.
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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+2*a*ds
vf^2=(20m/s)^2+2*(980cm/s^2)*120cm
vf=sqrt((20m/s)^2+2*(980cm/s^2)*120cm)
vf=485.39m/s
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That's cm/s, as I expect you see.
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ds=120cm
dv=465.39m/s
vAve=(485.39m/s+20m/s)/2=505.39/2=252.7
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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?
vf=v0+a*dt
485.39cm/s=20cm/s+980cm/s^2*dt
465.39cm/s=980cm/s^2*dt
dt=.47s
v0=80cm/s
dt=.47s
ds=80cm/s*.47s=37.6cm
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Very good but see my notes and be careful to distinguish between m/s and cm/s.
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