Phy 231
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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: If the ball returned to the same position it was release from it had to have a horizontal motion of 10 m/s for .5 s, which gives us 5 m.
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: the motion would be similar to a curved parabolic arc due to the combined motion of being thrown upwards and the horizontal motion equal to that of the vehicle.
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: As mentioned above the path would be that of a parabolic arc, which can be further described as opening downwards.
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: The vertical velocity from moment of release would be equal to that of vf = v0 + a 'dt, where the final velocity after .25 s is the apex of the vertical motion; therefore, 0 = v0 – 9.8 m/s^2 (.25 s), which is 2.45 m/s = v0. This gives us a velocity from the perspective of an observer of v = 'sqrt [( 2.45 m/s)^2 + (10 m/s)^2] = 'sqrt ( 106.0025) = 10.30 m/s
How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion: The ball will rise a distance of 'ds = 2.45 m/s ( .25 s) = 0.6125 m
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15 minutes
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Solution
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