course phy121
I added comments to this. Your comments are marked with &&...&&The comments I added are marked with &&&&...&&&&
cq_1_182
phy121
Your 'cq_1_18.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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?
10m/s * .5sec = 5m
As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
Due to the ball, passenger, and child all traveling at the same speed as the vehicle, the ball would appear to have a perfectly verticle path.
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?
It would be a semicircle or arch with a 5m base with a peak at rougly .306m.
&&Good answer. It would actually be a parabola, but a parabolic arc remains nearly congruent for a ways with a circular arc and it would require mathematical analysis beyond the scope of your course to prove the difference.&&
&&&&I had originaly written parabola, but I removed it because I didn't know if a parabola needed to have a closed base or if it could be just an ""arc"" for lack of a better term.&&&&
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?
The ball will be moving 2.45m/s on the y-axis.
The ball would appear to have a velocity of (2.45m/s theta 90*) + (10m/s theta 0*) = 10.3m/s theta 13.77*
The velocity at this instant would be the result of a vertical 2.45 m/s velocity and a 10 m/s horizontal velocity.
&&This result would be a vector of length sqrt( (2.45 m/s)^2 + (10 m/s)^2) = 10.2 m/s, approx., at angle theta = arcTan(2.45 m/s / (10 m/s) ) = arcTan(.245) = 16 degrees, give or take the degree or so.&&
&&&& I've run both tan^-1(2.45/10) and tan^-1(.245) through my TI89 and got 13.7663 both times. I'm not trying to be rude or anything, but how am I missing this?&&&&
How high did the ball rise above its point of release before it began to fall back down?
Your notation
(2.45m/s theta 90*) + (10m/s theta 0*) = 10.3m/s
wasn't clear. You didn't indicate how you got the angle. So my note indicated the details, just to be sure.
13.776 rounds to 14, which is within 'a unit or so' of 16 degrees. However my 16 degree estimate wasn't particularly accurate.
.306M
I determined this looking at only the downward side of the arch. Using `ds = (vF / 2) * `dt where vF = 2.45m/s and `dt is .25sec.
See my notes. It seems clear that you understand this.
&#Good responses. See my notes and let me know if you have questions. &#"