Physics Question (In Description)

Question

A fireperson, 50.0 m away from a burning building, directs a stream of water from a fire hose at an angle of 30° above the horizontal. If the velocity of the stream is 40.0 m/s, at what velocity will the stream of water strike the building?

Answer 1

This question is asking us to find the final velocity when water strikes the building (I'll call it "v_f"). We are given the following info:

v_i = 40.0 m/s

θ = 30 degrees above horizontal

Δx = 50 m

First, we separate the overall initial velocity into its horizontal and vertical components, using trig functions.

v_ix = v_i*cosθ = (40.0 m/s)*cos(30 deg) = 34.64 m/s

v_iy = v_i*sinθ = (40.0 m/s)*sin(30 deg) = 20.00 m/s

Now, in typical projectile motion under ideal conditions, there are a few things that happen. The horizontal component of velocity remains constant throughout the entire flight. Also, the vertical component changes as gravity acts on the projectile. We can look at this as two separate motions superimposed upon each other to get an idea of the overall projectile motion.

Let's start with the easier component:

v_fx = v_ix = 34.64 m/s

We can also calculate the total flight time of the projectile, since we know the horizontal distance traveled and the constant horizontal velocity at which we travel.

t = Δx/v_x = (50 m)/(34.64 m/s) = 1.44 sec

Now, for the vertical component:

a = -9.81 m/s² (constant acceleration due to gravity)

v_fy = v_iy + at = 20 m/s + (-9.81 m/s²)*(1.44 sec) = 5.87 m/s

v_fy = 5.87 m/s

We now have our final horizontal velocity and final vertical velocity. The last step is to combine them to find the total velocity.

v_f = sqrt[(v_fx)² + (v_fy)²] = sqrt[(34.64 m/s)² + (5.87 m/s)²]

v_f = 35.13 m/s, (since v_fy was positive this implies water from hose is still traveling upwards when it hits the building, it has not yet reached its peak)