phy121
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
• At what clock time(s) will the speed of the ball be 5 meters / second?
• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion:
The object reaches its maximum height when vf = 0, therefore average velocity will be ‘dv/2, or 7.5 m/s.
To get ‘ds, we can use
Vf^2 = v0^2 + 2 * a * ‘ds
0^2 = (15 m/s)^2 + (2* -10 m/s^2 * ‘ds)
0 = 225 m^2/s^2 + -20 m/s^2 * ‘ds
-225 m^2/s^2 = -20 m/s^2 * ‘ds
-225 m^2/s^2 / -20 m/s^2 = ‘ds
11.25 m = ‘ds above initial height, or 11.25 m + 12 m = 23.25 m total height.
If we know that it travels 11.25 m at an average velocity of 7.5 m/s, we can determine the time it takes to travel that distance by using 11.25 m / 7.5 m/s = ‘dt . The time it took, therefore, would be 1.5 s.
We know that the velocity is 0 m/s when the object is 23.25 m above the ground. Acceleration will be positive after this point because the object is falling with gravity.
Vf^2 = v0^2 + 2 * a * ‘ds
Vf^2 = 0 + 2* 10 m/s^2 *23.25 m
Vf^2 = 465 m^2/s^2
Vf = 21.56 m/s when object hits the ground.
Since we know it took 1.5 s to reach the max height, we need to determine how long it took to hit the ground, then add the two together.
vAve = (0+ 21.56 m/s) / 2
vAve = 10.78 m/s
‘ds / vAve = ‘dt
23.25 m / 10.78 m/s = 2.16 s
2.16 s + 1.5 s = 3.66 s
To determine at what clock time the velocity will be 5 m/s, we use aAve = ‘dv/’dt.
-10 m/s^2 = (5 m/s – 15 m/s) / ‘dt and 10 m/s^2 = (5– 0 m/s) / ‘dt
-10 m/s^2 = -10 m/s/’dt 10 m/s^2 = 5 m/s / ‘dt
-10 m/s^2 * ‘dt = -10 m/s 10 m/s ^ 2/ 5 m/s = ‘dt
‘dt = 1 sec. ‘dt = 2 sec.
To determine when the object will be 20 m. off the ground, we use ‘ds = v0 * ‘dt + .5 a * ‘dt
20 m = 15 m/s * ‘dt + .5 * -10 m/s^2 * ‘dt and 20 m = 0 m/s * ‘dt + .5 * 10 m/s^2 * ‘dt
20 m = ‘dt (15 m/s + -5 m/s^2) 20 m = 5 m/s^2 * dt
20 m / (15 m/s – 5 m/s^2) = ‘dt 20 m / 5 m/s^2 = ‘dt
(20 m / 15 m/s) – (20m / 5 m/s^2) = ‘dt sqrt 4 s^2 = ‘dt
4/3 s – 4 s^2 = ‘dt ‘dt = 2 sec
4/3 s – 2 s =’dt
.67 s = ‘dt
6 seconds – 1.5 seconds = 4.5 seconds after object reaches maximum height.
V0 * ‘dt + .5 a * ‘dt = ‘ds
0 * 4.5s + .5 (10 m/s^2) * 4.5 s = ‘ds
0 + (5 m/s^2 * 4.5 s) = ‘ds
‘ds = 22.5 meters from maximum height, so 23.25 m – 22.5 m = .75 m above the ground.
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30 minutes
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Your work is excellent, but be sure to see the link below, and be sure you understand how the final velocity can be found just from the initial velocity, acceleration and net displacement, with no need to break the motion into two phases or to reverse your positive direction. No need to submit a revision unless you have a question.
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#