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

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Homework Help Overview

The discussion revolves around a 2D kinematics problem involving projectile motion, specifically calculating the correct launch angle for a package based on its vertical and horizontal motion components. The original poster presents calculations that yield an angle of 31 degrees, while a reference suggests 42 degrees.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods for determining the angle, including using vertical and horizontal velocity components, distances, and time of flight. Questions arise regarding the appropriateness of using final velocities versus distances for the arctangent calculation.

Discussion Status

Multiple interpretations of the problem are being explored, with participants offering different approaches to the calculations. Some guidance has been provided regarding the use of horizontal velocity and the significance of time in the calculations. There is no explicit consensus on the correct method or outcome yet.

Contextual Notes

Participants note potential issues with rounding errors and the precision of calculations, as well as the implications of constant horizontal velocity in the context of the problem.

deserthobo
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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:
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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
 
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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.
 
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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
 
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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
 

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