A gun is fired from a moving platform and ranges of the shot are observed to be $R_{1}$ and $R_{2}$ when the platform is moving forwards and backwards, respectively, with velocity $v_{P}$ . Find the elevation of the gun $α$ in terms of the given quantities.

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Solution

Horizontal range $= \frac{2 u_{x} u_{y}}{g}$
Let the initial horizontal and vertical components of the velocity of the shot be $u$ and $v$ , respectively, w.r.t. platform.
$R_{1} = \frac{2 (u + v_{P}) v}{g}$ , platform moving forward
$R_{2} = \frac{2 (u - v_{P}) V}{g}$ , platform moving backward
$R_{1} + R_{2} = \frac{4 u v}{g}$
and $R_{1} - R_{2} = \frac{4 v_{P} v}{g}$
Now $(R_{1} - R_{2})^{2} = \frac{16 v_{p}^{2} v^{2}}{g^{2}}$
and $\frac{(R_{1} - R_{2})^{2}}{R_{1} + R_{2}} = \frac{4 v_{P}^{2}}{g} \times \frac{v}{u}$
or $\frac{v}{u} = \frac{g}{4 v_{p}^{2}} \frac{(R_{1} - R_{2})^{2}}{(R_{1} + R_{2})}$
Elevation of gun
$α = {tan}^{- 1} (\frac{v}{u}) = {tan}^{- 1} [\frac{g}{4 v_{p}^{2}} \times \frac{(R_{1} - R_{2})^{2}}{(R_{1} + R_{2})}]$ .

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A gun is fired from a moving platform and ranges of the shot are observed to be R1 and R2 when the platform is moving forwards and backwards, respectively, with velocity vP. Find the elevation of the gun α in terms of the given quantities.

A gun is fired from a moving platform and ranges of the shot are observed to be $R_{1}$ and $R_{2}$ when the platform is moving forwards and backwards, respectively, with velocity $v_{P}$ . Find the elevation of the gun $α$ in terms of the given quantities.