over level ground, the greatest range of a gun is 25km. the muzzle velocity of a bullet leaving the barrel of this gun is ?
no idea where to start.
Well, the greatest range is achieved on level ground when a projectile is fired at an angle of 45 degrees with the horizontal.
I don't know if you're supposed to know the range formula,. So, I'll use the long way.
Let the muzzle velocity be v.
Along the horizontal, you have a component $\displaystyle v \cos(45)$
Along the vertical, you have a component $\displaystyle v \sin(45)$
The time for the projectile to land is obtained using the vertical component:
$\displaystyle s = ut + \dfrac12 at^2$
s = 0,
u = v cos(45)
t = ?
a = -9.8
From this, you get an equation with v and t.
The next part is when taking the horizontal component.
In time t, it covers distance d.
$\displaystyle d = ut$
d = 25 km = 25000 m
u = v sin(45)
t = ?
This gives you another equation with v and t.
Solve simultaneously to get the value of v.