How Do You Calculate Max Relative Error in Projectile Motion?

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SUMMARY

The calculation of max relative error in projectile motion is determined using the formula s = (v^2sin2t)/g, where g is the acceleration due to gravity (9.81 m/s²). Given a launch angle of 30 degrees and relative errors of 10% for both initial velocity (v) and angle (t), the derived formula for relative error is ds/s = 2dv/v + 2cot(2t)dt. By substituting the values, the max relative error in horizontal distance (s) is conclusively calculated to be 40%.

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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.
 
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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%
 

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