Grade 11General Physics

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

Ananya Sharma

10 Years agoGrade 11

2 Answers

Akshay

Approved Tutor Answer10 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

8 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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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

Ananya Sharma

2 Answers

Akshay

Anjali yadav

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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

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the earth’s gravitational pull (its weight) but deviates slightly from this direction. Calculate the
deviation (a) at 40 latitude, (b) at the poles, and (c) at the equator