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Assignment 18 Seed Question 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: The horizontal displacement of the ball should be zero, since the ball itself is moving fast enough and for short enough time that the child can still catch the ball while the car is still moving. In other words, the ball was thrown up in a straight fashion and caught once it returned to the child’s hand, so it should not have moved any.
• 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: As observed, the path of the ball to a passenger in the car would appear to follow a straighet, vertical path up and then 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: The sketch looks like a ball going up and then coming down in an arc type fashion. So, the shape looks like an arc.
• 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: At the instant of release, we can find how fast the ball was moving by knowing that the vf = 0 m/s at its highest point and knowing the equation is equal to 5 m = .5 sec * (0 m/s – v0) / 2. We can then solve for v0 as 20 m/s. The velocity, as observed by the people on the side of the road will be less quantitative, as they will just see the ball go up, appear to stand still for a moment, and then go down once again.
The only thing related to 5 meters appears to be the 10 m/s horizontal velocity. However this velocity is constant, and is not in any case relevant to the vertical motion.
What do you know about the vertical motion of the ball for this interval? What can you therefore conclude?
• How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: The height of the ball’s rise would be calculated by ‘ds = ‘dt * vAve = 10 m/s * .5 sec = 5 m. This is before it falls back down again.
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I spent about 25 minutes on this assignment.
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Very good up to a point. Then you appear to have confused vertical and horizontal quantities into the same equation.
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