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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?
v0_ver=20cm/s, s0_ver=0cm, a_ver=980cm/s^2
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• What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?
vf_ver=sqrt(v0_ver^2 + 2*a*’ds)=sqrt((20cm/s)^2+2*980cm/s^2*120cm)=485cm/s
‘ds_ver=sf_ver - s0_ver=120cm - 0cm=120cm
‘dv_ver=(vf_ver - v0_ver)=(485cm/s - 20cm/s)=465cm/s
vAve_ver=( vf_ver + v0_ver)/2=(485cm/s + 20cm/s)/2=253cm/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?
‘dt_hor=’dt_ver=sqrt((‘ds-v0)/(.5*a))=sqrt((120cm-20cm/s)/(.5*980cm/s^2))=.452s
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One of your calculations involves subtracting v0 from `ds.
`ds - v0 is not a meaningful calculation. You can't subtract a velocity from a displacement.
Put another way, `ds is a number and a unit, and the unit is a unit of length. v0 is a number and a unit, and its unit is of length divided by time. The two are therefore unlike terms and cannot be added or subtracted.
Specifically, 120 cm - 20 cm/s is not a meaningful calculation.
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The problem is in the algebra of solving the fourth equation for v0. Redo that solution without looking at what you did before, and be careful.
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v0_hor=80cm/s
a_hor=(vf_hor-v0_hor)/’dt=(0cm/s - 80cm/s)/.452s=-177m/s^2
????I had to assume that vf_hor=0cm/s to find accel. I’m not sure how to find accel with the only horizontal info I had which was ‘dt=.452s and v0_hor=80cm/s. I only had two parts of info. I need 3 to solve for the accel. How should I solve for accel hor at this point????
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For an ideal projectile, there is zero force in the horizontal direction.
So you know `dt, v0 and a for the horizontal motion.
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