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course Phy 121
re-submit
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:
‘dt = 0.5 second
a = 9.8 m/s^2
v = 10 m/s
‘ds = (10 m/s + 0 m/s)/2 * 0.5 s
‘ds = 2.5 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:
The ball would appear to move in an arc. The ball would go up and begin to curve toward the back of the car.
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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:
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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:
a = -9.8 m/s^2
vf = 0 m/s
dt = 0.5 s
vf = v0 + a ‘dt
0 m/s = v0 - 9.8 m/s^2 * 0.5 s
4.9 m/s = v0
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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:
‘ds = (vf + v0)/2 * ‘dt
‘ds = 1.225 m
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There is no need for a revision.
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Solution
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