91Ó°ÊÓ

Launch angle is 45^o
retarding force =${\stackrel{→}{F}}_{\text{wind}}=-F_{\text{wind}}\hat{i}$

mass = m

The flight duration in projectile motion is

$T=\frac{2v_{0}sin\left(\mathrm{θ}\right)}{g}...................................\left(1\right)$

The displacement s of an object moving with acceleration and initial speed u is given by

$s=ut+\frac{1}{2}a{t}^2....................................\left(2\right)$

The vertical component of the projectile is not affected by the horizontal retarding force of the wind.
Find the horizontal range by substituting T from equation (1) in equation 2

$$\begin{gathered} R=v_{0}cos\left(\mathrm{θ}\right)T-\frac{1}{2}a{T}^2 \\ R=v_0\cos \left(\mathrm{θ}\right)\left(\frac{2v_0\sin \left(\mathrm{θ}\right)}{g}\right)-\frac{1}{2}a{\left(\frac{2v_0\sin \left(\mathrm{θ}\right)}{g}\right)}^2 \\ R=\left(\frac{{v_0}^2\sin \left(2\mathrm{θ}\right)}{g}\right)-\left(\frac{2a{v_0}^2{\sin }^2\left(\mathrm{θ}\right)}{g}\right) \end{gathered}$$

Range is maximum when

$\frac{dr}{d\mathrm{θ}}=0$

differentiate

$$\begin{gathered} \frac{d}{d\mathrm{θ}}\left[\frac{v_0^2\sin \left(2\mathrm{θ}\right)}{g}-\frac{2a{v}_0^2{\sin }^2\left(\mathrm{θ}\right)}{g^2}\right]=0âˆ\pounds \\ \frac{2v_0^2\cos \left(2\mathrm{θ}\right)}{g}-\frac{2a{v}_0^2\sin \left(2\mathrm{θ}\right)}{g^2}=0 \\ \frac{\cos \left(2\mathrm{θ}\right)}{g}=\frac{asin\left(2\mathrm{θ}\right)}{g^2} \\ \tan \left(2\mathrm{θ}\right)=\frac{g}{a}…………………………………\left(3\right) \end{gathered}$$

retardation is a= F_{wind /}m, substitute this in equation (3)

$$\begin{gathered} \tan \left(2\mathrm{θ}\right)=\frac{mg}{F_{\text{wind}}} \\ 2\mathrm{θ}={\tan }^{-1}\left(\frac{mg}{F_{\text{wind}}}\right)âˆ\pounds \\ \mathrm{θ}=\frac{1}{2}{\tan }^{-1}\left(\frac{mg}{F_{\text{wind}}}\right) \end{gathered}$$

Mass = m

initial speed = v₀

horizontal retarding force =${\stackrel{→}{F}}_{\text{wind}}=-F_{\text{wind}}\hat{i}$

Angle of projection is 45^o

When there is no wind , the retardation due to resistance is zero.

The maximum range without wind is

$R==\frac{v_0^2\sin (2×45°)}{g}=\frac{v_0^2}{g}$

The maximum range with the wind is given as

$$\begin{gathered} R_w=\frac{v_0^2\sin \left(2\mathrm{θ}\right)}{g}-\frac{2a{v}_0^2{\sin }^2\left(\mathrm{θ}\right)}{g^2} \\ {R}_w=\frac{v_0^2\sin \left(2\mathrm{θ}\right)}{g}-\frac{2\left(\frac{F_{\text{wind}}}{m}\right)v_0^2{\sin }^2\left(\mathrm{θ}\right)}{g^2} \\ {R}_w=\frac{v_0^2}{g^2}\left(g\sin \left(2\mathrm{θ}\right)-\frac{2F_{\text{wind}}{\sin }^2\left(\mathrm{θ}\right)}{m}\right) \\ \mathrm{θ}=\frac{1}{2}{\tan }^{-1}\left(\frac{mg}{F_{\text{wind}}}\right) \\ \mathrm{θ}=\frac{1}{2}{\tan }^{-1}\left(\frac{0.5×9.81}{0.6}\right) \\ \mathrm{θ}=41.513° \\ {R}_w=\frac{v_0^2}{g}\left(\sin \left(2\mathrm{θ}\right)-\frac{2F_{\text{wind}}{\sin }^2\left(\mathrm{θ}\right)}{mg}\right) \\ {R}_w=\frac{v_0^2}{g}\left(\sin \left(2×41.513°\right)-\frac{2×0.6{\sin }^2\left(41.513°\right)}{0.5×9.81}\right) \\ {R}_w=\frac{v_0^2}{g}(0.885 \end{gathered}$$

Now find the percent decrease

$$\begin{gathered} \frac{R-R_w}{R}×100\%=\frac{\frac{v_0^2}{g}-\frac{v_0^2}{g}(0.885)}{\frac{v_0^2}{g}}×100\% \\ \frac{R-R_w}{R}×100\%=(1-0.885)100\% \\ \frac{R-R_w}{R}×100\%=11.49\% \end{gathered}$$