Pistol Signal Flare Range: Utah & Moon (55° Angle)

[Fanjoni]

A pistol that fires a signal flare gives the flare an initial speed (muzzle speed) of 185 m/s.
(a) If the flare is fired at an angle of 55° above the horizontal on the level salt flats of Utah, what is its horizontal range? You can ignore air resistance.
(b) If the flare is fired at the same angle over the flat Sea of Tranquility on the moon, where g = 1.6 m/s2, what is its horizontal range?

I am just going blank i tried solving for range but am lost

 

[OlderDan]

A pistol that fires a signal flare gives the flare an initial speed (muzzle speed) of 185 m/s.
(a) If the flare is fired at an angle of 55° above the horizontal on the level salt flats of Utah, what is its horizontal range? You can ignore air resistance.
(b) If the flare is fired at the same angle over the flat Sea of Tranquility on the moon, where g = 1.6 m/s2, what is its horizontal range?

I am just going blank i tried solving for range but am lost

Resolve the initial velocity into horizontal and vertical components. Use what you know about motion with constant acceleration to figure out how long it will take for an object moving upward with the vertical component of the flare velocity to return to the ground. The horizontal motion of the flare has no effect on the vertical motion, so the time in flight depnds only on the initial vertical velocity and the acceleration. Once you have the time in flight, use the (constant) horizontal component of velocity to figure out how far the flare moves horizontally; that is the range.

 

[Fanjoni]

Thanks i finaly got it it took me a while
This is what i did

185*sin55 =V(ertical speed)
V/g = T(ime to stop vertically) T*2=total time
185*cos55 = v(horizontal speed)

t*v=R(horizontal range)

I did the same for the moon g = 1.6

Thanks