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Grade: 12
        A machine gun is mounted on the top of a tower 100 m 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 150m/s,take g=10m/s^2
2 years ago

Answers : (1)

Arun
23389 Points
							
Dear Astha
 

Here, 
u = 150 ms-1
And, Taking g = 10 $ms^{-2}$ (approximately) 
Let 'θ' be the angle at which the machine gun should fire in order to cover a maximum distance. 
Then , the horizontal component of velocity = 150 cos θ 
And, the vertical component of velocity = 150 sin θ 
If 'T' is the time of flight, then , 
Horizontal range, R = (150 cos θ ) x T . 
The gun is mounted at the top of a tower 100 meters high . 
Let us regard the positive direction of the position-axis as to be along the line from the top of tower in downward direction. 
For motion along vertical :
Initial Velocity = -150 sin θ; 
Distance covered = + 100 m 
And, acceleration = + 10 $ms^{-2}$
In time 'T' , the machine gun shot will reach maximum height and then reach the ground. 
Now, 
S = ut + 1/2 a$t^{2}$
Therefore, +100 = ( -150 sin θ ) T + 1/2x10x$T^{2}$
Or, $T^{2}$- ( 30 sin θ )T - 20 = 0 
Therefore, T = - ( - 30 sin θ ) ± { ( - 30 sin θ)2 - 4 x 1 x (-20) }1/2/2 

= 30 sin θ ± (900 sin2 θ + 80 )1/2/2 
Or, T = 15 sinθ ± (225 sin2θ + 20)1/2
Now, range will be maximum, if time of flight is maximum . 
Therefore, choosing positive sign ,we have 
T = 15 sin θ + ( 225 sin2θ+ 20 )1/2
Hence, horizontal range covered, 
R = 150 cos θ {15 sin θ+ ( 225 sin2θ + 20 )}1/2
The horizontal range is maximum, when θ = 45o. 
But in the present case , the machine gun is mounted at height of 100 m  
Therefore, R will not be maximum for θ = 45o.It will be maximum for some value of θ close to 45
If we calculate values of R by setting θ = 43, 43.5, 44, 45, 46 and 47, the values of R come out to be 2347 m, 2347.7 m, 2348 m, 2346 m, 2341 m and 2334 m respectively. 
Thus R is maximum for value of θ some where between 43.5 and 44. 
Therefore , the mean value of θ = (43.5 + 44)/2 = 43.75. 
The gun should be inclined at 43.75 to cover a maximum range of firing on the ground below. 

 

Regards

Arun (askIITians forum expert)

2 years ago
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