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A gun can fire shells with maximum speed $v_{0}$ and the maximum horizontal range that can be achieved is $R=v{_{0}}^{2}/g$

If a target farther away $\Delta x$ beyond R has to be hit with the same gun as shown in the figure. Show that it could be achieved by rating the gun to a height at least

$h=\Delta x[1+\frac{\Delta x}{R}]$

Answers (1)

The solution to this problem can be given in two ways:

i) The target is present at the horizontal distance of $(R+\Delta x)$ and is h meters below the projection point. $Y=-h$

ii) The motion of the projectile starting from point P till reaching point T. The vertical height covered is -h and the horizontal range is $\Delta x$

max range of a projectile is given by $R=\frac{v^{2}}{g}$

here, $\theta =45$

we assume that the gun is raised to a height of h to hit the target T

total range $=(R+\Delta x)$

the horizontal component of velocity is, $v\; \cos \theta$

horizontal velocity at $P=V_{x}=-V\; \cos \theta$

vertical velocity at $P=V_{y}=V\sin \theta$

now, $h=ut+\frac{1}{2}at^{2}$

So, $h=-V\; \sin \theta t+\frac{1}{2}gt^{2}\; \; \; \; \; \; ---------------(1)$

$V\; \cos \theta .t=(R+\Delta x)$

$t=\frac{(R+\Delta x)}{V\; \cos \theta }$

Substituting the value of t in (1) we get,

$h=-V\; \sin \theta \left ( \frac{(R+\Delta x)}{V\; \cos \theta } \right )+\frac{1}{2}g\left ( \frac{(R+\Delta x)}{V\; \cos \theta } \right )^{2}$

$h=-(R+\Delta x)+\frac{1}{R}(R^{2}+\Delta x^{2}+2R\; \Delta x)=-R-\Delta x+R+\frac{\Delta x^{2}}{R}+2\Delta x$

$h=\Delta x[1+\frac{\Delta x}{R}]$

hence proved.

Posted by

infoexpert23

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