Question

The range in meters of a projectile launched over a flat ground from the origin with positive velocity $V$ in $m/s$ at an angle $\theta$ given in radian is given by $R=\frac{V^2\sin \left(2\theta \right)}{g}$ where $g$ is a positive constant, assume $V=2m/s,g=10m/s^2$ and $\theta$ was measured to be $\frac{\pi }{12}$ radians. If there was a possible error in the measurement of $\theta$ of $\frac{1}{10\sqrt{3}}$ radians, estimate the corrosponding error in the computation of the range.

• A. $0.04m$
• B. $0.4m$
• C. $0.08m$
• D. $0.8m$

Answer 1

The correct option is A $0.04m$

$f(x+\Delta x)=f\left(x\right)+f^{{}^{\prime }}\left(x\right).\Delta x$
Range $=R\left(\theta \right)=\frac{v^2}{g}\sin 2\theta$
$R(\theta +\Delta \theta )=R\left(\theta \right)+R^{{}^{\prime }}\left(\theta \right).\Delta \theta$
Error $=\Delta R=R^{{}^{\prime }}\left(\theta \right).\Delta \theta$
$\Delta \theta =\frac{1}{10\sqrt{3}},R\left(\theta \right)=\frac{v^2}{g}\sin 2\theta \Rightarrow R^{{}^{\prime }}\left(\theta \right)=\frac{v^2}{g}\times 2\cos 2\theta$
$=\frac{4}{10}\times 2\times \frac{\sqrt{3}}{2}=\frac{4\sqrt{3}}{10}$
$\Delta R=\frac{4\sqrt{3}}{10}\times \frac{1}{10\sqrt{3}}=\frac{4}{100}=0.04\text{m}$