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The maximum range of a gun on horizontal surface is 16km . If g=10 ms -2 , the muzzle velocity of the shell must be:- a. 1600 ms -1 b. 400 ms -1 c. 200 ( root2 ) ms -1 d. 160(root 10) ms -1 ​I want the answer with full solution

The maximum range of a gun on horizontal surface is 16km. If g=10 ms-2, the muzzle velocity of the shell must be:-
a. 1600 ms-1          b. 400 ms-1           c. 200 (root2 ) ms-1         d. 160(root 10) ms-1

​I want the answer with full solution

Grade:11

2 Answers

Akshay
185 Points
8 years ago
maximum range will be when gun is shot at 45 deg.
Lets prove this fact first, assume that gun is shot at angle a with horizontal with velocity v,
U = (v.cosa) i + (v.sina) j
s=U*t+0.5*a*t2 ,
a= – 10 j,
s = (v.cosa*t) i + (v.sina*t – 5.t2) j,
final y-coordinate will be 0,
so , t=v.sina / 5,
Range = v.cosa.t = v2.cosa . sina / 5 = v2 . sin2a /10,
maximum range will be when sin2a = 1, or a=45,
Max range =  v2 / 10 = 16000,
So v=400 m/s
Anjali yadav
13 Points
5 years ago
The horizontal range will be maximum if the bullet is fired at an angle of 45° to the horizontal. We have, R=u^2*sin2β/g Since the angle is 45°,soR=u^2*(sin2*45°)/gor, R=u^2/gAlso, R=16km=16000mAnd, g=10m/s^2So, 16000=u^2/10or, u^2=16000*10or, u=√160000Therefore, u=400m/s Hope you understand this😊😊

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