Mrak-Face
26-12-2007, 09:10 AM
I had to do a short, simple responsive answer into the classical physics component of an everyday activity in my life. I chose something that happens at parties I attent (Called CC).
I just need you to note if I made any errors here.
The following a physics based concept, around a popular event that normally occurs at CC gatherings, I suggest you read it, as it may be of great importance to the next time you decide to go drunken jumping.
Alison is at a party which is being held at Nicoles house. She is drunk and bored and decides to do some jumping. She then finds the small ledge in Nicoles backyard which is exactly 1.5m high and decides to jump off of it. Alison jumps off of the ledge with an initial velocity of 4.5ms-1 at an angle of 36.7 degrees.
With this information we can find Alisons Horizontal and Vertical velocity. Y dignifys horizontal displacements, X dignifys vertical displacements.
Uy = 4.5Cos36.7
Uy = 3.607ms-1
Ux = 4.5sin36.7
Ux = 2.689ms-1
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram1.jpg (http://photobucket.com/)
With this we can calculate the maximum height of Alison with the following information and equation of projectile motion:
Ux = 2.689ms-1
V = 0ms-1 (Due to irregular gravitational stance of 0g)
Ax = -9.8ms-2 (Gravity acting as a negative to Alison)
Sx = ?
Therefore we can use the following equation of projectile motion:
V^2 = u^2 - 2as
: 0 = (2.689^2) - 2*-9.8ms-2*?
: 7.2307 = -19.8*?
We must arrange the equation to find ? (Sx)
: ?(Sx) = 7.2307/19.6 = 0.3689m
Therefore, Alison jumped 36.89cm, with an additional 1.5m from the ledge, she reached a maximum height of 1.8689m.
Now with this height, we can calculate the time Alison spent in air.
V = 0ms-1 (Calculating from 0g nav point)
Ax = -9.8ms-2 (The gravity acting upon Alison)
Ux = 2.689ms-1
T = ?
With this gathered information we can use the following rule of projectile motion:
v = u + at
To find ?(T) we must re-arrange the equation to get the following.
?(t) = v - ux / ax
: ?(t) = (0ms-1 - 2.689ms-1 / -9.8ms-1)
:: ?(t) = 0.274387755s
::: t = 0.274387755s
We can then plot this information onto the running graph.
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram2.jpg (http://photobucket.com/)
From this displacement of 1.8689m we can calculate the GPE of Alison and therefore work from to find the total time spent in midair.
GPE = mgh
M = 55kg
G = 9.8ms-2
H = 1.8689m
GPE = 55*9.8ms-2*1.8689
Alison's GPE from her highest point is 1007.3371J
Along with this we can finish her total time spent in air using the following information.
S = 9.8ms-1 (Acting constant from 0.274387755s of freefall of Alison)
D = 1.8689m
T = ?
S = d/?(t)
We must arrange the equation to get ?(t), therefore:
?(t) = d/s
: ?(t) = 1.8689m/9.8ms-1
:: ?(t) = 0.1907s
With this, in addition to the 0.274387755s and in addition to this the 0.1907s of freefall from the total max height we can gather that Alison has a total fall time of : 0.465087755s, and a total air time of 0.739475510s.
We can now plot this onto the running graph.
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram3.jpg (http://photobucket.com/)
Finally, with this information we can calculate the total distance Alison would travel, we can use the following information.
Uy = 3.607ms-1
Ay = 0ms-1 (Not including wind resistance)
T = 0.739475510s
Sy = ?
With this we can use the following equation of projectile motion:
S = ut + 1/2at^2
: ?(Sy) = 3.607ms-1*0.739475519s + 1/2*0ms-1*0.739475510s^2
:: ?(Sy) = 2.66728819m
With this information we can complete our running graph:
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram4.jpg (http://photobucket.com/)
: If Alison were to jump off of Nicole's 1.5m ledge at a party of hers, while drunk, at a speed of 4.5ms-1 and an angle of 36.7 degrees, she would spend 0.739475510s in air, 0.465087755s of that being freefall. She would travel 0.3689m vertically and 2.66728819m horizontally and finally, she would hit the ground with a force of 1007.3371J, which is equivilant to dropping a 3lt carton of milk off of a 8 story building.
Next time, I'll go into the circular motion of pinging.
I just need you to note if I made any errors here.
The following a physics based concept, around a popular event that normally occurs at CC gatherings, I suggest you read it, as it may be of great importance to the next time you decide to go drunken jumping.
Alison is at a party which is being held at Nicoles house. She is drunk and bored and decides to do some jumping. She then finds the small ledge in Nicoles backyard which is exactly 1.5m high and decides to jump off of it. Alison jumps off of the ledge with an initial velocity of 4.5ms-1 at an angle of 36.7 degrees.
With this information we can find Alisons Horizontal and Vertical velocity. Y dignifys horizontal displacements, X dignifys vertical displacements.
Uy = 4.5Cos36.7
Uy = 3.607ms-1
Ux = 4.5sin36.7
Ux = 2.689ms-1
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram1.jpg (http://photobucket.com/)
With this we can calculate the maximum height of Alison with the following information and equation of projectile motion:
Ux = 2.689ms-1
V = 0ms-1 (Due to irregular gravitational stance of 0g)
Ax = -9.8ms-2 (Gravity acting as a negative to Alison)
Sx = ?
Therefore we can use the following equation of projectile motion:
V^2 = u^2 - 2as
: 0 = (2.689^2) - 2*-9.8ms-2*?
: 7.2307 = -19.8*?
We must arrange the equation to find ? (Sx)
: ?(Sx) = 7.2307/19.6 = 0.3689m
Therefore, Alison jumped 36.89cm, with an additional 1.5m from the ledge, she reached a maximum height of 1.8689m.
Now with this height, we can calculate the time Alison spent in air.
V = 0ms-1 (Calculating from 0g nav point)
Ax = -9.8ms-2 (The gravity acting upon Alison)
Ux = 2.689ms-1
T = ?
With this gathered information we can use the following rule of projectile motion:
v = u + at
To find ?(T) we must re-arrange the equation to get the following.
?(t) = v - ux / ax
: ?(t) = (0ms-1 - 2.689ms-1 / -9.8ms-1)
:: ?(t) = 0.274387755s
::: t = 0.274387755s
We can then plot this information onto the running graph.
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram2.jpg (http://photobucket.com/)
From this displacement of 1.8689m we can calculate the GPE of Alison and therefore work from to find the total time spent in midair.
GPE = mgh
M = 55kg
G = 9.8ms-2
H = 1.8689m
GPE = 55*9.8ms-2*1.8689
Alison's GPE from her highest point is 1007.3371J
Along with this we can finish her total time spent in air using the following information.
S = 9.8ms-1 (Acting constant from 0.274387755s of freefall of Alison)
D = 1.8689m
T = ?
S = d/?(t)
We must arrange the equation to get ?(t), therefore:
?(t) = d/s
: ?(t) = 1.8689m/9.8ms-1
:: ?(t) = 0.1907s
With this, in addition to the 0.274387755s and in addition to this the 0.1907s of freefall from the total max height we can gather that Alison has a total fall time of : 0.465087755s, and a total air time of 0.739475510s.
We can now plot this onto the running graph.
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram3.jpg (http://photobucket.com/)
Finally, with this information we can calculate the total distance Alison would travel, we can use the following information.
Uy = 3.607ms-1
Ay = 0ms-1 (Not including wind resistance)
T = 0.739475510s
Sy = ?
With this we can use the following equation of projectile motion:
S = ut + 1/2at^2
: ?(Sy) = 3.607ms-1*0.739475519s + 1/2*0ms-1*0.739475510s^2
:: ?(Sy) = 2.66728819m
With this information we can complete our running graph:
http://i121.photobucket.com/albums/o238/MarkTheBetter/Diagram4.jpg (http://photobucket.com/)
: If Alison were to jump off of Nicole's 1.5m ledge at a party of hers, while drunk, at a speed of 4.5ms-1 and an angle of 36.7 degrees, she would spend 0.739475510s in air, 0.465087755s of that being freefall. She would travel 0.3689m vertically and 2.66728819m horizontally and finally, she would hit the ground with a force of 1007.3371J, which is equivilant to dropping a 3lt carton of milk off of a 8 story building.
Next time, I'll go into the circular motion of pinging.