Phy 201
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A child in a car tosses a ball upward so that after release it requires 1/2 second to rise and fall back into the child's hand at the same height from which it was released. The car is traveling at a constant speed of 10 meters / second in the horizontal direction.
• Between release and catch, how far did the ball travel in the horizontal direction?
answer/question/discussion: ->->->->->->->->->->->-> :
10m/s * .5 s
= 5m
good
#$&*
• As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
answer/question/discussion: ->->->->->->->->->->->-> :
Straight up and down
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• Sketch the path of the ball as observed by a line of people standing along the side of the road. Describe your sketch. What was shape of the path of the ball?
answer/question/discussion: ->->->->->->->->->->->-> :
Ball went up at a forward motion and back down
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• How fast was the ball moving in the vertical direction at the instant of release? At that instant, what is its velocity as observed by a line of people standing along the side of the road?
answer/question/discussion: ->->->->->->->->->->->-> :
5m * .5s = 2.5m/s
5 m * .5 s is 2.5 m * s; this isn't a velocity.
The ball rises for 1/4 second, at which time its velocity is zero. It is accelerating downward at 9.8 m/s^2.
Thus you know vf, a and `dt for the rising ball. You can find `ds and v0.
#$&*
• How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: ->->->->->->->->->->->-> :
#$
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15 min
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