Phy 121
Your 'cq_1_08.2' 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 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 max height is 11.25 m after 1.5 sec. When it hits the ground it will be going 21.56 m/s. This will occur 3.656 seconds after the initial toss. The speed of the ball will be 5 m/s at clock time = 1 sec and 2 sec. I think the ball will be 20 meters above the ground at clock times = 0.694 s and 2.306. I think after 6 second the ball will be 18.25 meters above the ground unless I got lost in all of this.
v0 = 15 m/s
a = -10 m/s^2
vf = 0 m/s
vf = v0 + a * `dt
`dt = (vf – v0) / a
`dt = ( 0 m/s – 15 m/s) / -10 m/s^2
`dt = -15 m/s / -10 m/s^2 = 1.5 s
`ds = (v0 + vf) / 2 * `dt
`ds = (15 m/s + 0 m/s) / 2 * 1.5 s
`ds = 7.5 m/s * 1.5 s
`ds = 11.25 m
Ground level equal 11.25 m + 12 m = when `ds = 23.25 m, so from max. height to ground level:
Vf^2 = v0^2 + 2a`ds
Vf^2 = (0 m/s)^2 + 2 * 10 m/s^2 * 23.25 m
Vf^2 = 0 + 465 m^2/s^2
Vf^2 = 465 m^2/s^2
Vf = 21.56 m/s
Vf = v0 + a * `dt
21.56 m/s = 0 m/s + 10 m/s^2 * `dt
21.56 m/s = 10 m/s^2 * `dt
(21.56 m/s) / (10 m/s^2) = `dt
`dt = 2.156 s
Time from initial toss to hitting ground = 1.5 s + 2.156 s = 3.656 s
At what clock time will the speed be 5 m/s:
On the way up:
Vf = v0 + a * `dt
5 m/s = 15 m/s + -10 m/s^2 * `dt
-10 m/s = -10 m/s^2 * `dt
-10 m/s / -10 m/s^2 = `dt
`dt = 1 sec
It will also be traveling 5 m/s on the way down:
Vf = v0 + a * `dt
5 m/s = 0 m/s + 10 m/s^2 * `dt
5 m/s = 10 m/s^2 * `dt
5 m/s / 10 m/s^2 = `dt
`dt = .5 s which is clock time 2 s (1.5 + .5)
It will be 20 meters above the ground, 8 meters above the toss on the way up:
`ds = v0`dt + .5a`dt^2
8 m = 15 m/s `dt + .5(-10 m/s^2)`dt^2
8 m = 15 m/s`dt -5 m/s^2`dt^2
0 = -5 m/s^2`dt^2 + 15 m/s`dt – 8 m
I’m not sure I can use this here because the units are different but,
`dt = [-b +-sqrt(b^2 – 4ac)] / 2a
`dt = [-15 +- sqrt(15^2 – 4*-5*-8)] / (2 * -5)
you didn't include the units here but they would work out
`dt = [ -15 +- sqrt65] / -10
`dt = (-15 +-8.06) / -10
`dt = (-15 + 8.06) / -10 and (-15 – 8.06) / -10
`dt = -6.94 / -10 and -23.06 / -10
`dt = .694 and 2.306
On the way down it will be 20 meters above the ground 3.25 meters after begins to fall (23.25 – 20):
`ds = v0`dt + .5a`dt^2
3.25 m = 0`dt + .5(10 m/s^2)`dt ^2
0 = 5 m/s^2`dt^2 -3.25 m
`dt = [-b +- sqrt(b2 – 4ac)] / 2a
`dt = +- sqrt (0 – 4 * 5 * -3.25) / 2* 5
`dt = +- sqrt 65 / 10
`dt = +-8.06 / 10
`dt = .806 and -8.06 Clock time .806 + 1.5 = 2.306
It takes 3.656 seconds to reach the ground so it will then begin going back up. I’m not sure it will be at the same velocity with which it hit the ground, but:
V0 = 21.56 m/s
Vf = 0 m/s
A = -10 m/s^2
Time to reach max height again:
Vf = v0 + a * `dt
0 m/s = 21.56 m/s + -10 m/s^2 * `dt
0 = 21.56 m/s – 10m/s^2`dt
-21.56 m/s = -10 m/s^2`dt
`dt = -21.56 m/s / -10 m/s^2
`dt = 2.156 second or clock time 5.812 (2.156 + 3.656)
`ds = (v0 + vf)/2 *`dt
`ds = 10.78 m/s * 2.156 s
`ds = 23.24 m so max height the second time will be 23.24 m above the ground
It will then begin to fall again and we want to know how high it will be .188 second after it begins to do so:
V0 = 0 m/s
Vf = ?
`ds = ?
`dt = .188 s
A = 10 m/s^2
Vf = v0 + a * `dt
Vf = 0 m/s + 10 m/s^2 * .188 s
Vf = 1.88 m/s
`ds = (v0 + vf) / 2 * `dt
`ds = (0 m/s + 1.88 m/s) / 2 * .188 s
`ds = .94 m/s * .188 s
`ds = 5 m so it will have fallen 5 meters from the max height of 23.25 meters above the ground so it will be 18.25 meters above the ground (23.25 – 5) at clock time equal 6 sec.
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30 minutes
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Almost everything you did is very good. However you should see the link for a couple of insights (e.g., you don't always have to break the motion into separate 'up' and 'down' phases):
&#Please compare your solutions with the expanded discussion at the link
Solution
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