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Over level ground, the greatest range of a gun is 25 km. The muzzle velocity of a bullet leaving the barrel of this gun is:

A) 495 ms^-1

B) 510 ms^-1

C) 525 ms^-1

C) 480 ms^-1

A) 495 ms^-1

B) 510 ms^-1

C) 525 ms^-1

C) 480 ms^-1

so initial displacement vertically and horizontally are both zero, right?

The net initial velocity is the one we need to find out. We don't know the angle nor do we know the maximum points nor do we know... anything. How to do this? The wording is not very clear to me.

I tried substituting $x = ut\cos\theta , y = ut\sin\theta - 4.9t^2$ to eliminate time but it brings useless quadratic equations which have funknown factors

I know that v = 0 when $x = 12500$, but that's as much as I can figure.

Any good hints would be appreciated.

does it say that there is any wind resistance?

if not. the greatest range is when the projectile is aimed at 45degrees to the horizontal.

if not. the greatest range is when the projectile is aimed at 45degrees to the horizontal.

Vjyrik

so initial displacement vertically and horizontally are both zero, right?

The net initial velocity is the one we need to find out. We don't know the angle nor do we know the maximum points nor do we know... anything. How to do this? The wording is not very clear to me.

I tried substituting $x = ut\cos\theta , y = ut\sin\theta - 4.9t^2$ to eliminate time but it brings useless quadratic equations which have funknown factors

I know that v = 0 when $x = 12500$, but that's as much as I can figure.

Any good hints would be appreciated.

The net initial velocity is the one we need to find out. We don't know the angle nor do we know the maximum points nor do we know... anything. How to do this? The wording is not very clear to me.

I tried substituting $x = ut\cos\theta , y = ut\sin\theta - 4.9t^2$ to eliminate time but it brings useless quadratic equations which have funknown factors

I know that v = 0 when $x = 12500$, but that's as much as I can figure.

Any good hints would be appreciated.

Time of flight is [2usin*]/g and so range = [u^2][sin2*]/g.

This range is max when * = 45 degrees. Hence max range = [u^2]/g. Etc.

Original post by Maths Buster

Time of flight is [2usin*]/g and so range = [u^2][sin2*]/g.

This range is max when * = 45 degrees. Hence max range = [u^2]/g. Etc.

This range is max when * = 45 degrees. Hence max range = [u^2]/g. Etc.

how did you get these equations??

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