Question

A machine gun is mounted on top of a tower 100m high.At what angle should the gun be inclined to cover max range . The muzzle speed of the bullett is 150m/s

Answer 1

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. 

Answer 2

Answer:

Explanation:

Here in this question the given data ,

Height of the tower = h= 100 m,

Muzzle speed of bullet = u = 150 m/s

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

Then the horizontal component of velocity = 150 cosθ

and the vertical component of velocity = 150 sinθ

If T is the time of flight then ,horizontal range ,R = (150 cosθ)×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 sinθ

Distance covered = 100 m

Acceleration due to gravity = g= 10 m/s2

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

Now from laws of motion ,

S= ut +12at2⇒100=(−150sinθ)×T +12×10×T2⇒T2−(30 sinθ)×T −20 =0⇒T=30sinθ±12(900sin2θ+80)2⇒T =15sinθ±(225sin2θ+20)

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

Therefore positive sign we have ,

T =15sinθ+(225sin2θ+20)

Hence horizontal range covered,

R=150cosθ×[15sinθ+(225sin2θ+20)]

Now the horizontal range is maximum when θ=45°

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

So at θ=45° the range will not be maximum , it will be maximum for some value of θ which are close to 450 .

Now if we calculate the values of R by setting θ=43°,43⋅5°,44°,45°,46°,47° the values of R come out to be 2347 m, 2347⋅7m,2348m, 2346m,2341m,2334m

Thus R is maximum for some value θ between 43° and 43⋅5° .

Therefore the mean value of θ=43+43⋅52=43⋅75° .

Therefore the gun should be inclined at θ=43⋅75° to cover maximum range firing on the ground below .

Regards