What is the correct angle for the given 2D kinematics problem?

AI Thread Summary
The discussion revolves around solving a 2D kinematics problem involving a package's trajectory with a resultant velocity of 240 m/s and a vertical distance of 2400 m. The user calculated vertical and horizontal velocity components, determining the time for the package to fall and subsequently the horizontal distance traveled. Confusion arose regarding the correct method to calculate the angle, with initial attempts yielding 31 degrees, then 62 degrees, and finally 43 degrees, while the book states 42 degrees. The final consensus suggests that rounding errors may account for the discrepancy, and the importance of using the correct velocities in calculations is emphasized. Ultimately, the discussion highlights the complexities of kinematic equations and the potential for minor errors in calculations.
deserthobo
Messages
13
Reaction score
0

Homework Statement


http://img413.imageshack.us/img413/2919/physicsprobpr6.jpg

Xy=2400 m

Homework Equations



Xy=Voyt + 1/2at^2
Xx=1/2(Vox+Vx)t

The Attempt at a Solution



I calculated the vertical and horizontal velocity components of the given resultant velocity of 240 m/s.

Voy = 120 m/s
Vox = 208 m/s

then i used the vertical component, the given vertical distance and gravity to calculate the time needed for the package to fall to the ground.

t = 38 s

then i plugged this into Xx=1/2(Vox+Vx)t to find the horizontal distance the package travelled.

Xx= 3904

then i plugged this in with the given vertical distance into tan -1 (y/x) to get an angle of 31 degrees. but the book says 42 degrees! where did i go wrong guys?
 
Last edited by a moderator:
Physics news on Phys.org
you want the arc tan of the final velocities Vyf/Vxf
 
hmm why do i want the arc tan of the final velocities and not the arc tan of the distances?

also, arent the final velocites both = 0 since the package hits the ground
 
Last edited:
denverdoc, I think deserthobo is right is the arctan of distances, the 2.4 km and the horizontal range of the flare.
 
hmm...the time needed for the object to fallwell, vf = vo + at

vf = 0 so...
 
Last edited:
ok using vf=vo + at i got 12 s

i plugged 12 s into x = 1/2(vf + vo)t and then found the arc tan with that number and came up with 62 degrees. doesn't work out.
 
Last edited:
ok guys, if i plug in the horizontal velocity component which is 208 for

vf and vo

for

x=1/2(vf+vo)t

i will get 2704 m, which when plugged into arc tan (y/x) will give me the angle of 41..5 which when rounded up is 42 degrees. but does it make sense to plug in 208 for both vf and vo?
 
ok, t is not just 12, there are some extra digits as well (i'm horrible with sig digs as well :-P)

also, the horizontal velocity is constant

and velocity = distance/time

so, distance = velocity * time

you know the time, and you know the velocity
 
Last edited:
ok so i did everything following what mybsaccownt said and in the end i came up with 43 degrees. the book says 42 so this method is incorrect!
 
  • #10
deserthobo said:
ok so i did everything following what mybsaccownt said and in the end i came up with 43 degrees. the book says 42 so this method is incorrect!
I think you and the book are likely bothright, round off error. And i was all wet, in a hurry between patients , and shouldn't have posted.
 
  • #11
Are you a doctor? :)
 
  • #12
guilty as charged, tho a psychiatrist in the minds of many of his fellow physicians has lost any valid claim to the title
 
Back
Top