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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? (15m/s)^2 / 2a = (15m/s)^2 / 2(-10m/s/s) = 11.25m + 12m above the ground = 23.25m above the ground
· How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? Don’t know
You know v0, a and `ds:
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
Between the initial instant and striking the ground the ball has a displacement of 12 meters downward. This is the displacement from the initial position to the ground. So `ds = -12 m.
Which equations of motion apply when you know v0, a and `ds?
· At what clock time(s) will the speed of the ball be 5 meters / second? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
The relevant interval is the interval between the initial instant and the instant the ball's velocity reaches 5 m/s. On this interval the final speed of the ball is 5 m/s, so its velocity is either 5 m/s or -5 m/s.
In either case you know v0, vf and a.
How do you reason out a situation in which you know v0, vf and a?
How do the equations of uniformly accelerated motion apply to a situation where you know v0, vf and a?
Now you have two choices for vf. Consider each choice separately:
Assuming vf = 5 m/s, you can find `dt and `ds. What do you get?
Assuming vf = -5 m/s, you can find `dt and `ds. What do you get?
· At what clock time(s) will the ball be 20 meters above the ground? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
For the interval of interest here, the ball will displace +8 m from the initial point to its highest point. So `ds = 8 m.
So you know v0, a and `ds. How do you solve a situation in which you know v0, a and `ds?
What do you get in this case?
· How high will it be at the end of the sixth second? Don’t know
Assuming upward to be the positive direction, v0 = 15 m/s and the acceleration (being downward, in the direction opposite to the chosen positive direction) is a = -10 m/s^2.
For this interval `dt = 6 seconds.
So you know v0, a and `dt. How do you solve a situation in which you know v0, a and `dt?
What do you get in this case?
answer/question/discussion:
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i spent about 1 hr on this seed question and the previous one.
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I'm lost. I tried the book but I didn't get far.
For each of the above situations you know v0 and a. Different situations imply different intervals, so in some cases you know vf, in some you know `dt, in some you know vf. Putting any of these three together with v0 and a you have enough information to either solve the equations for or reason out all other quantities.
See my notes.
Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&&.