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.
Between release and catch, how far did the ball travel in the horizontal direction?
answer/question/discussion:
Because of the constant speed .5*10m/s=5m/s
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:
It went straight up then fell straight back 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:
It started from his hand and the ball went up and to the right to its peak then fell back
down to the right. Shaped like a half circle
Good description; in fact it's a parabola, but a 'shallow' parabola has a shape very close to that of the top of a circle.
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:
Since we have a distance of .3m to its highest point and it does it in .25 sec then
vAve=`ds/`dt which is .3m/.25sec, vAve=1.2m/s. Since at its highest point v will be 0 then
intial v must be double the vAve which 1.2m/s*2=v0=2.4m/s
How high did the ball rise above its point of release before it began to fall back down?
answer/question/discussion:
Since we know `dt and a we can find `ds.Since a =9.8m/s^2 and at the point of its highest
point is v0=0, and since we are calculating the distance down it will take a .25 sec we get
`ds=vo`dt+.5a`dt^2,ds=.5(9.8m/s^2).25sec^2,`ds=4.9m/s^2*.06scs,`ds=.3m
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20min
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&#Good work. See my notes and let me know if you have questions. &#