Results 1 to 12 of 12
Thread: More Physics

20080907, 11:22 #1
 Join Date
 Feb 2002
 Location
 A Magic Forest in Deepest, Darkest Kent
 Posts
 5,681
 Thanks
 0
 Thanked 1 Time in 1 Post
Re: More Physics
<P ID="edit" class=small>(Edited by Jezza on 07Sep08 23:22. pressed submit by mistake)</P>Hi Jordan,
Is it vectors homework this week <img src=/S/grin.gif border=0 alt=grin width=15 height=15>.
The simplest way to approach these questions/puzzles is to draw a triangle. If the kayaka travels north a distance of 3 metres, she is pushed 2 metres east.
Draw a triangle with a 2 inch horizontal base and a vertical line 3 inches long perpendicular and join making a tringle. The Hypotenuse is therefore
sqrt(2^2+3^2)= 3.6 metres
Therefore for every 3 metres she travels north she is travelling 3.6 metres. Therfore the total distanvce she travels is 100/3 * 3.6=120 metres
Now using the trigonometry you can find the angle she travels.
The other answers are the same, just draw trianglesJerry

20080907, 11:26 #2
 Join Date
 Mar 2001
 Location
 Dallas, Texas, USA
 Posts
 1,680
 Thanks
 0
 Thanked 1 Time in 1 Post
Re: More Physics
If you think you have the solution for Q4, you should post what you have. Would probably be better for someone to confirm your solution (or to show where you went wrong), than to put forth the effort required to simply duplicate your answer.
<img src=/S/2cents.gif border=0 alt=2cents width=15 height=15> Ricky

20080907, 11:49 #3
 Join Date
 Jun 2002
 Location
 Mt Macedon, Victoria, Australia
 Posts
 3,993
 Thanks
 1
 Thanked 45 Times in 44 Posts
Re: More Physics
Q5 The package is dropped from a height of 100m, and falls under gravity. Use the stuff from the previous set of questions to work out how long it will take to reach the ground.
The plain travels at constant speed. Work out how far it will travel in the time you just worked out.
This distance is the answer to the question, because the package keeeps moving forward with the speed of the plain while it falls.Regards
John

20080907, 11:59 #4
 Join Date
 Jun 2002
 Location
 Mt Macedon, Victoria, Australia
 Posts
 3,993
 Thanks
 1
 Thanked 45 Times in 44 Posts
Re: More Physics
There is a standard formula for the range of a projectile. Have you been given that?
The max occurs when the angle is 45 degrees , because sin(90) is 1, giving a max of v^2/g
So if the range at some other angle is half the max, it equals v^2/2g.
Use the formula and solve of the angle. (The V and g are on both sides so will drop out.)Regards
John

20080907, 12:18 #5
 Join Date
 Aug 2008
 Location
 St. Louis, Missouri, USA
 Posts
 0
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
My solution for Q4:
Horizontal velocity = 20m/s*cos5 degrees = 19.92m/s
Initial vertical velocity = 20m/s*sin5 degrees = 1.74m/s
Initial vertical position = 2.0m
Use the equation v = s / t to find out how long it takes the ball to get to the net:
t = s / v = (7.0m) / (19.92m/s)
t = 0.35s
Use the equation s = Vit + .5at^2 to find the vertical change in position of the ball:
s = (1.74m/s)(0.35s) + .5(9.81m/s^2)(0.35s)^2
s = 0.612m  0.605m
s = 0.007m
The height of the ball when it reaches the net is 2.0m  0.007m = 1.993m
Since the net is 1.0m high, the ball cleared the net by 0.993m
This answer just doesn't seem right, so I was hoping someone could show me what I did wrong (or tell me that I did it right <img src=/S/joy.gif border=0 alt=joy width=23 height=23>).
I also worked out Q5...
Initial vertical position = 100m
Initial vertical velocity = 0m/s
Horizontal velocity = 150m/s
I used the equation s = Vit + .5gt^2 to find the amount of time it would take to fall:
Vi = 0, so s = .5gt^2
t = sqrt[(2s)/g]
t = sqrt[ (2*100m)/(9.81m/s^2) ]
t = 20.39s
Then I used v = s / t to find how far it would travel horizontally in 20.39s.
s = vt
s = (150m/s)(20.39s)
s = 3,058m

20080907, 12:23 #6
 Join Date
 Aug 2008
 Location
 St. Louis, Missouri, USA
 Posts
 0
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
Thank you! <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

20080907, 13:47 #7
 Join Date
 Aug 2008
 Location
 St. Louis, Missouri, USA
 Posts
 0
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
I haven't been given a formula.
Sorry, I don't understand...

20080907, 14:23 #8
 Join Date
 Aug 2008
 Location
 St. Louis, Missouri, USA
 Posts
 0
 Thanks
 0
 Thanked 0 Times in 0 Posts
More Physics
I'm very confused. I really appreciate any help.
Q1. A kayaker needs to paddle north across a 100m wide harbor. The tide is going out, creating a tidal current that flows to the east at 2.0m/s. The kayaker can paddle with a speed of 3.0m/s.
Which direction should he paddle to travel straight across the harbor?
How long will it take him to cross?
Q2. A ball is thrown at 20m/s. It leaves the thrower's hand 1.8m above the ground and is caught at the same height 30m away. What is the maximum height of the ball on its way to the catcher?
I don't know how to do this b/c I can't find the initial horizontal or vertical velocity of the ball...
Q3. At what launch angles will a projectile land at half of its maximum possible range on flat ground? (This one is giving me the most trouble...) <img src=/S/brickwall.gif border=0 alt=brickwall width=25 height=15>
Q4. A tennis player hits a ball 2.0m above the ground. The ball leaves his racquet with a speed of 20m/s at 5.0 degrees above the horizontal. The horizontal distance to the net is 7.0m, and the net is 1.0m high. Does the ball clear the net? By how much does it clear or miss?
I think I've figured this one out, but I want to see if anyone else gets the same answer.
Q5. A supply plane needs to drop a package. The plane flies 100m above the ground at a speed of 150m/s. How far short of the target should it drop the package?

20080908, 03:26 #9
 Join Date
 Nov 2001
 Location
 Melbourne, Victoria, Australia
 Posts
 5,016
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
Q2. A ball is thrown at 20m/s. It leaves the thrower's hand 1.8m above the ground and is caught at the same height 30m away. What is the maximum height of the ball on its way to the catcher?
I don't know how to do this b/c I can't find the initial horizontal or vertical velocity of the ball...
This problem is, indeed, ambiguous but still solvable. It does seem to be drawing on analysis that (seems to me) is beyond the kind of stuff you’ve been doing in the other problems. It's quite a "tuff" problem.
There are only two trajectories that can pass through two given points. For this problem, make the two points (0,0) and (30,0) – here 0 for the ycoordinate is at the 1.8m hand height. We know the initial speed u is 20m/s, but don’t know the angle alpha. The following is one solution, starting with the xy cartesian equation for a parabolic trajectory. Hope it helps rather than confuses.
Alan

20080908, 11:21 #10
 Join Date
 Jun 2002
 Location
 Mt Macedon, Victoria, Australia
 Posts
 3,993
 Thanks
 1
 Thanked 45 Times in 44 Posts
Re: More Physics
Does this make sense?
Regards
John

20080908, 13:41 #11
 Join Date
 Nov 2001
 Location
 Melbourne, Victoria, Australia
 Posts
 5,016
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
Perhaps I should add here that it was not necessary to find the actual value(s) of the angle to answer the question. Using g=10m/s²:

20080909, 15:53 #12
 Join Date
 Aug 2008
 Location
 St. Louis, Missouri, USA
 Posts
 0
 Thanks
 0
 Thanked 0 Times in 0 Posts
Re: More Physics
Thank you Alan!