A machine gun is mounted on the top...

A machine gun is mounted on the topp of the tower 100m high at what on should the gun is inclined to cover the manimum range of firing on the the speed of bullet is 150m/s (g = 10 m/s) ground below. The go

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Height of the tower ( =mathrm{h}=100 mathrm{m} ) Muzzle speed of bullet ( =mathrm{u}=150 mathrm{m} / mathrm{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 ) cos ( theta ) and the vertical component of velocity ( =150 ) sin ( theta ) If ( mathrm{T} ) is the time of flight then , horizontal range ( , mathrm{R}=(150 cos theta) times T ) Now as the gun is mounted on the tower of ( 100 mathrm{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 sin theta ) Distance covered ( =100 mathrm{m} ) Acceleration due to gravity ( =g=10 mathrm{m} / mathrm{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=u t+frac{1}{2} a t^{2} ) ( Rightarrow 100=(-150 sin theta) times T+frac{1}{2} times 10 times T^{2} ) ( Rightarrow T^{2}-(30 sin theta) times T-20=0 ) ( Rightarrow T=frac{30 sin theta pm frac{1}{2}(900 sin 2 theta+80)}{2} ) ( Rightarrow T=15 sin theta pm(225 sin 2 theta+20) ) Now the range will be maximum if the flight time is maximum . Therefore positive sign we have, ( T=15 sin theta+(225 sin 2 theta+20) ) Hence horizontal range covered, ( R=150 cos theta times[15 sin theta+(225 sin 2 theta+20)] ) Now the horizontal range is maximum when ( theta=45^{circ} ) But in this case the machine gun is mounted at the height of ( 100 mathrm{m} ) So at ( theta=45^{circ} ) the range will not be maximum, it will be maximum for some value of ( theta ) which are close to ( 45^{circ} ) Now if we calculate the values of R by setting ( theta=43^{circ}, 43 cdot 5^{circ}, 44^{circ}, 45^{circ}, 46^{circ}, 47^{circ} ) the values of R come out to be ( 2347 mathrm{m}, 2347 cdot 7 mathrm{m}, 2348 mathrm{m}, 2346 mathrm{m}, 2341 mathrm{m}, 2334 mathrm{m} ) Thus ( mathrm{R} ) is maximum for some value ( theta ) between ( 43^{circ} ) and ( 43 cdot 5^{circ} ). Therefore the mean value of ( theta=frac{43+43-5}{2}=43 cdot 75^{circ} ) Therefore the gun should be inclined at ( theta=43 cdot 75^{circ} ) to cover maximum range firing on the ground below
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