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A machine gun is mounted on the top of a tower 100m high. At what Angle should the gun be inclined to cover a maximum range of firing on the ground below? The muzzle speed of the bullet is 150ms^{-1}.

Here in this question the given data ,

Height of the tower = h= 100 m,

Muzzle speed of bullet = u = 150 m/s

Let $\theta $ be the angle at which machine gun would fire in order to cover maximum distance .

Then the horizontal component of velocity = $150\mathrm{cos}\theta $

and the vertical component of velocity = $150\mathrm{sin}\theta $

If T is the time of flight then ,horizontal range ,R = $\left(150\mathrm{cos}\theta \right)\times T$

Now as the gun is mounted on the tower of 100 m high .

Therefore, the positive direction of the position axis as to be along the line from the top of the tower in downward direction .

For motion along vertical :

Initial velocity = $-150\mathrm{sin}\theta $

Distance covered = 100 m

Acceleration due to gravity = g= 10 m/s

^{2}

Now in time T the machine gun shot will reach maximum height and then reach the ground .

Now from laws of motion ,

$S=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 100=\left(-150\mathrm{sin}\theta \right)\times T+\frac{1}{2}\times 10\times {T}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {T}^{2}-\left(30\mathrm{sin}\theta \right)\times T-20=0\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{30\mathrm{sin}\theta \pm \frac{1}{2}\left(900\mathrm{sin}2\theta +80\right)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow T=15\mathrm{sin}\theta \pm \left(225\mathrm{sin}2\theta +20\right)\phantom{\rule{0ex}{0ex}}$

Now the range will be maximum if the flight time is maximum .

Therefore positive sign we have ,

$T=15\mathrm{sin}\theta +\left(225\mathrm{sin}2\theta +20\right)$

Hence horizontal range covered,

^{$R=150\mathrm{cos}\theta \times \left[15\mathrm{sin}\theta +\left(225\mathrm{sin}2\theta +20\right)\right]$}

Now the horizontal range is maximum when $\theta =45\xb0$

But in this case the machine gun is mounted at the height of 100 m

So at $\theta =45\xb0$ the range will not be maximum , it will be maximum for some value of $\theta $ which are close to 45

^{0}.

Now if we calculate the values of R by setting $\theta =43\xb0,43\xb75\xb0,44\xb0,45\xb0,46\xb0,47\xb0$ the values of R come out to be $2347m,2347\xb77m,2348m,2346m,2341m,2334m$

Thus R is maximum for some value $\theta $ between $43\xb0and43\xb75\xb0$ .

Therefore the mean value of $\theta =\frac{43+43\xb75}{2}=43\xb775\xb0$ .

Therefore the gun should be inclined at $\theta =43\xb775\xb0$ to cover maximum range firing on the ground below .

Regards

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