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.
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Between release and catch, how far did the ball travel in the horizontal direction?
answer/question/discussion: 10 m/s * .5 s = 5 m
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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, straight 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: parabola
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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: vy = 1/2(9.8 m/s^2)(.5 s) = 4.9 m/s; 10 m/s
The ball is moving in the y direction at 4.9 m/s, and in the x direction at 10 m/s. Its velocity at that instant is therefore found by the Pythagorean Theorem to be sqrt( (4.9 m/s)^2 + (10 m/s)^2 ) = sqrt(124 m^2/s^2) = 11.2 m/s, approx.
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How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: y = 1/2gt^2 = 4.9 m/s^2 * (.5 s)^2 = 1.225 m
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10 minutes
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&#Good responses. See my notes and let me know if you have questions. &#