Water leaves the end of a fire hose at a velocity of 14m/s. If the hose is held at an angle of 30degrees from?

Question

the horizontal, how far will the water reach neglecting air resistance?
treat the the water as u would an individual object.
(hint: use your vector skills to break the velocity into its two components: the amount of velocity acting straight upwards will ditermine how long the object is in the air, the componient acting, the componient acting horizontally wiill ditermine how far it travels in that time)

Answer 1

The horizontal speed is Cos(30)*14
and the vertical speed is
sin(30)*14
assuming the fire hose is at ground level, then the water will rise and then fall according to
y=sin(30)*14*t-.5*g*t^2
y=0 at t=0 and
t=2*sin(30)*14/g
plug the t into the horizontal equation of motion
x=cos(30)*14*t
x=.5*cos(30)*g/sin(30)

j

Answer 2

Distance = 2 (v sinθ / g) (v cosθ) = v²/g sin(2θ)

Answer: L = v²/g sin(2θ) = 17.3 m