How Do You Calculate Max Relative Error in Projectile Motion?

  • Thread starter Thread starter D44
  • Start date Start date
  • Tags Tags
    Differentiation
D44
Messages
70
Reaction score
0
I'm trying to solve the problem but keep getting the wrong answer. Could someone help please?

s = (v^2sin2t)/g

where s = horizontal distance, v = initial velocity, t = angle, g = acc. gravity (9.81 m/s).

Intended launch angle is 30 degrees, find max relative error in s if v and t are subject to relative errors of 10%.

I've got:

ds/dv = (2vsin2t)/g
ds/dt = (2v^2cos2t)/g

ds/s = 2dv/v + 2cot2t.dt

How do I calculate this if I don't have a value of v?

I kinda got to another step (which is still wrong I think), which is...

ds/s = 2v/10v + (2cot(2t).dt)/10 where I thought maybe the v's cancelled.

Any help would be appreciated.
 
Physics news on Phys.org
ds/s = 2dv/v + 2(cos2t/sin2t)dt
If you consider here sin2t=x,then (cos2t dt/sin2t)=dx/x
Hence ds/s=2dv/v+2dx/x
substituting dv/v=10 and dx/x=10
ds/s=40%.
Hence relative error of s is 40%
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top