Answer:
0.05
Explanation:
Given a variable C which is the product of two variables A, B:
[tex]C=A\cdot B[/tex]
Then the absolute error on C is given by:
[tex]\frac{\sigma_C}{C}=\frac{\sigma_A}{A}+\frac{\sigma_B}{B}[/tex]
where [tex]\sigma_A, \sigma_B, \sigma_C[/tex] are the uncertanties on the measure of A, B and C, respectively.
In this problem, the horizontal component of the velocity [tex]v_x[/tex] is given by
[tex]v_x = v_0 cos \theta[/tex]
Therefore, the uncertainty on vx is given by:
[tex]\frac{\sigma_{v_x}}{v_x}=\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta}[/tex] (1)
where we have:
[tex]v_0 = 7.94[/tex]
[tex]\sigma_{v_0}=0.03[/tex]
[tex]\theta=31^{\circ}[/tex], so
[tex]cos \theta=cos 31^{\circ}=0.857[/tex]
The uncertainty on [tex]cos \theta[/tex] is given by:
[tex]\sigma_{cos \theta}=|sin \theta|\sigma_\theta[/tex]
where:
[tex]|sin \theta| = |sin 31^{\circ}|=0.515[/tex]
and
[tex]\sigma_\theta=0.4^{\circ}=0.007 rad[/tex]
So
[tex]\sigma_{cos \theta}=|0.515|\cdot 0.007 = 0.0036[/tex]
Also,
[tex]v_x = v_0 cos \theta = (7.94)(cos 31^{\circ})=6.81 m/s[/tex]
So, combininb everything into (1), we find:
[tex]\sigma_{v_x}=(\frac{\sigma_{v_0}}{v_0}+\frac{\sigma_{cos \theta}}{cos \theta})v_x=(\frac{0.03}{7.94}+\frac{0.0036}{0.857})(6.80)=0.05[/tex]
