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.
** **
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?
it takes .5s to rise and a .5s to fall, total time is 1sec, therefore, 10m/s*1s = 10m
.5 sec is the time to rise and fall back, not just the time of rise or the time of fall. If it was the latter, though, your answer would be correct.
• As observed by a passenger in the car, what was the path of the ball from its release until the instant it was caught?
Straight up and straight 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?
I don’t know how to draw it. As you watch a car pass by, even if you could see the ball wouldn’t it appear to go straight up and down.
Relative to the roadside the ball travels at a constant horizontal velocity while rising more and more slowly, then falling more and more quickly. This gives it a curved path within that frame of reference. In fact it can (and will) be shown that the path would be parabolic, for the present a good description would indicate a curved path, rising or increasing at a decreasing rate then falling or decreasing at an increasing rate.
• 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?
v0 = vf - a`dt = 0 – (-9.8m/s^2)*.5s = 4.9m/s
• How high did the ball rise above its point of release before it began to fall back down?
y = (vf^2 – v0^2)/2a = -24.01m^2/s^2 / -19.6m/s^2 = 1.23m
** **
10mins
** **
You misread the time interval, so your solutions would need to adapt to the given time interval, but the procedure is correct and you could easily plug in the correct interval and get the correct results.
&#Let me know if you have questions. &#