phy 121
Your 'cq_1_18.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Seed question 18.2
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 * .5s = 5m
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 straight down.
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:
An arc with the beginning and ending pts being 5 m apart.
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:
I think I would need to know the distance it traveled up to figure out it’s vertical velocity. It wasn’t just dropped it was thrown up, but velocity depends on how high it went up and back down in the ½ second it traveled. However, at the top of the arc it stops and then starts to free fall again at an acceleration of 9.8m/s^2 for .25s
Aave * dt = dv
9.8m/s^2 * .25 = 2.45m/s
So the velocity ends at 2.45m/s so it must start at 2.45m/s and hit 0m/s at the top of the arc
How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion:
Ds = vAve*dt
VAve between release and top is 2.45/2 and dt is half of the total time (.25s)
Ds = (2.45/2) * .25
Ds = 1.225 * .25 = 0.31m
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20 mins
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