Vertical Projectile Motion Problem


by TheRedDevil18
Tags: motion, projectile, vertical
TheRedDevil18
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#1
Jan28-13, 01:04 PM
P: 153
1. The problem statement, all variables and given/known data

Please check my work.

Two students are on a balcony 19.6m above the ground. One guy throws a ball vertically downwards at 14.7m/s; at the same time the other guy throws a second ball vertically upwards at the same speed. The second ball just misses the balcony on the way down.

2.3.1) What is the difference in the balls flight time?
2.3.2) What is the velocity of each ball as they strike the ground?
2.3.3) How far apart are the balls 0.8s after they are thrown?

2. Relevant equations

vf^2 = vi^2 + 2a*y
delta y = vi*t + 1/2*a*t^2
delta y = vf+vi/2 * t
vf = vi + a*t

3. The attempt at a solution

2.3.1) For the ball thrown upwards I decided to first find the maximum height it reached. I used the first equation from the list and got an answer of 11.03m. I then used the third equation to find the time and got 1.5s (another 1.5s for coming down=3s). I then worked out the time for the ball thrown downwards to be 1s. My final answer for the difference in flight time is 4-1=3s.

2.3.2) Each ball would strike the ground at the same speed.
vf = vi+a*t
= 14.7+9.8*1
= 144.06m/s

2.3.3) Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.
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rude man
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#2
Jan28-13, 01:23 PM
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Quote Quote by TheRedDevil18 View Post
[b]1. 2.3.3) Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.
Check the number in red above! Otherwise all is good.
TheRedDevil18
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#3
Jan28-13, 01:33 PM
P: 153
I think the first ball should be 14.896 and the second ball 8.624. Is that my mistake?

rude man
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#4
Jan28-13, 01:40 PM
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Vertical Projectile Motion Problem


Quote Quote by TheRedDevil18 View Post
[b]1.

= 14.7+9.8*1 = 144.06m/s

Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.
Check the items in red above!
TheRedDevil18
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#5
Jan28-13, 01:46 PM
P: 153
Ok im a bit confused, but this is how I got 14.896m:
delta y = 14.7*0.8 + 1/2(9.8)*0.8^2
= 14.896m

Is there something I did wrong?
rude man
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#6
Jan28-13, 02:42 PM
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Quote Quote by TheRedDevil18 View Post
Ok im a bit confused, but this is how I got 14.896m:
delta y = 14.7*0.8 + 1/2(9.8)*0.8^2
= 14.896m

Is there something I did wrong?
Should that number be + or -?
TheRedDevil18
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#7
Jan29-13, 07:57 AM
P: 153
If the answer should be negative, Why?
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#8
Jan29-13, 10:43 AM
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Quote Quote by TheRedDevil18 View Post
If the answer should be negative, Why?
Because when you came up with y1 = +8.624m you defined y = 0 to be the balcony height, so anything below the balcony is negative. It should also be obvious that the second ball will be lower in height than the first, not higher as you had it.

So y1 = vi*t - gt^2/2 = 8.624m (ball tossed up but g is down)
y2 = -vi*t - gt^2/2 = -14.896m (ball tossed down & g is down)
y1 - y2 = 2vi*t = you fill this one in!

Just keep track of signs & you'll be fine.
TheRedDevil18
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#9
Jan29-13, 11:22 AM
P: 153
So the final answer is 23.52m apart?
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Jan29-13, 11:59 AM
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Quote Quote by TheRedDevil18 View Post
So the final answer is 23.52m apart?
Yep.
TheRedDevil18
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#11
Jan29-13, 12:21 PM
P: 153
Ok, thanks for your help


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