Your 'cq_1_8.2' 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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Seed quest 8.2
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 approxomation 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?
At the end of the 1st sec the v = 5m/s.
At the end of the 2nd sec the v = -5m/s
So, at the end of 1.5 sec the v = 0.
It takes 1.5 sec to get to it’s highest point.
ds = (v0 + vf)/2 * dt
ds = (15m/s + 0m/s)/2 * 1.5s
ds = 7.5m/s * 1.5 = 11.25m
It rises 11.25 m with a final position of 12m + 11.25m = 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?
This one took me a long time to figure out!
ds = -12m
a = -10 m/s/s
v0 = 15m/s
vf= ?
dt = ?
What is the velocity when it hit’s the ground?
vf^2 = v0^2 + 2a * ds
= 15^2 + 2(-10) * -12
= 225 + (-20 * -12)
= 225 + 240
vf^2 = 465
vf = sqrt 465 = 21.56
Except I think it is -21.56 because of the picture I drew and I take it to only 2 significant figures so vf = -22m/s.
There are two solutions to the equation vf^2 = 465 m^2 / s^2. The solutions are vf = +sqrt(465) m/s and vf = - sqrt(465) m/s, or approximately 21.6 m/s and -21.6 m/s.
If the upward direction is chosen as positive, then the negative solution is the one that corresponds to the conditions of the problem. The positive solution does not and it can be discarded.
Note the units calculation: sqrt(465 m^2 / s^2) = sqrt(465) sqrt(m^2/s^2) = sqrt(465) m/s.
How long after the initial toss?
Vf = v0 +a * dt
-22m/s = 15m/s + (-10m/s/s)*dt
-37= -10m/s/s*dt
3.7 = dt
The ball will hit the ground 3.7 s after it is tossed.
At what clock time(s) will the speed of the ball be 5 meters / second?
1 sec. For each second the velocity decreases 10m/s so if the initial vel is 15m/s then in one sec the velocity will be 5 m/s
The speed will also be 5 m/s when the velocity is - 5 m/s.
The latter occurs after 2 seconds.
Note that both intervals are associated with vf^2 = 25 m^2 / s^2. The fourth equation vf^2 = v0^2 + 2 a `ds yields only one solution for `ds, therefore telling us that both the positive and negative velocities are attained at the same vertical position.
At what clock time(s) will the ball be 20 meters above the ground?
The ball starts at 12m above the ground so we are trying to find the time it takes to get to ds = 8m.
ds = 8m
v0 = 15m/s
a = -10m/s/s
dt = ?
ds = v0 * dt + .5(a) * dt^2
I’m not sure if I can solve for dt in this equation and if I can, how. So I’ll find vf first using another equation and then solve for dt using another equation. Please explain if there is a simpler method.
vf^2 = v0^2 + 2(a) * ds
vf^2 = 15^2 + 2(-10) * 8
vf^2 = 225 -20*8
vf^2 = 225 - 160
vf^2 = 65
vf = sqrt 65 = 8.06 = 8
You would have positive and negative solutions, as in previous problems. In this case both the positive and negative solutions have meaning. The ball passes the 20 meter position going up and coming down.
Vf = v0 + a*dt
8m/s = 15m/s + (-10m/s/s)*dt
10m/s/s/*dt = 15m/s - 8m/s
Dt = 7m/s / 10m/s/s
Dt = .7s
The ball will be 20 m above the ground at 0.7s
How high will it be at the end of the sixth second?
answer/question/discussion:
Dt = 6s
V at 6s = 15m/s - 6(10m/s/s) = -45m/s
Vave = (15m/s - 45m/s)/2 = -15m/s
Ds = -15m/s * 6s = -90m
The ball started at 12m. The change in position at the end of 6 s is -90m so the final position at the end of 6s = 12 m - 90m = -78m.
Of course the ground will probably intervene, causing acceleration to become nonuniform and rendering this analysis meaningless.
On the other hand there could be a deep hole in the ground (e.g., a wellshaft).
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1.5 hours
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Very good work. See my notes, mostly about positive and negative solutions. The notes will show you what I mean.