A fireman d = 35.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 28.0° above the horizontal. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building? (Ignore the height of the hose nozzle above the ground, i.e. assume it is at ground level. Ignore air resistance in the motion of the water.)

Question

A fireman d = 35.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 28.0° above the horizontal. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building? (Ignore the height of the hose nozzle above the ground, i.e. assume it is at ground level. Ignore air resistance in the motion of the water.)

I tried to do tanθ = h/d, then I solved for h (height) but it turned out wrong. I'm not sure what to do with the given vi.

Answer 1

Vf^2=Vi^2+2ad

d=(Vf^2-Vi^2)/2a

Vf=0m/s^2
Vi=40m/s*Sin(28.0°)
a=-9.8m/s^2
d=????

Answer 2

break up the water stream into vertical, and horizontal components

Horizontal: vi=40cos28= 35.3 m/s
distance=30m=vihoriz*timein air
time in air= 30/35.3 seconds
Vertical: vi= 40sin28=18.8 m/s
hf=hi+vivertical*t-1/2 g t^2=0+18.8*30/35.3-4.9*(30/35.3)^2=....
which gives you the height the water strikes. I get about 12m high.
hf=hi+vi'*t-1/2 g t^2 where vi

Answer 3

40m/s*Cos(28.0°)= 35.32m/s

Vf=Vi. With projectile motion, there is no acceleration in the x-direction. So, Vf=Vi and t=D/Vx

35m/35.32m/s=t

Question asked how high up the wall, not how high the stream goes. Misread the question and used wrong kinematic equation and left out a step:

After solving for t using the horizontal velocity, use the vertical velocity to solve for D:

D=Vi*t+0.5at^2

A=-9.8m/s^2
Vi=40m/s*Sin(28.0°)
t=0.849s
d=????

Both setups should return the same answer.

Answer 4

Vo = 40m/s[28o].

Xo = 40*Cos28 = 35.3 m/s.
Yo = 40*sin28 = 18.8 m/s.

d = Xo*T = 35 m.
35.3 * T = 35,
T = 0.99 s. = Time in flight.
Tf = T/2 = 0.99/2 = 0.496 s. = Fall time.
h = 0.5g*Tf^2 = 4.9*0.496^2 = 1.21m.

Tr =

Answer 5

To solve this problem, we can break down the velocity of the water stream into its horizontal and vertical components.

First, let's find the horizontal component of the velocity (vix). We can use the given velocity magnitude (vi) and the angle above the horizontal (θi) to find it:

vix = vi * cos(θi)
vix = 40.0 m/s * cos(28.0°)

Next, let's find the vertical component of the velocity (viy). We can use the same given values to find it:

viy = vi * sin(θi)
viy = 40.0 m/s * sin(28.0°)

Now, we can analyze the horizontal and vertical motion of the water stream separately.

In the horizontal direction, there is no acceleration, so the horizontal velocity remains constant. The time it takes for the water stream to reach the building can be found using the horizontal distance (d = 35.0 m) and the horizontal velocity (vix).

d = vix * t
t = d / vix

Now, let's focus on the vertical motion. The water stream will follow a projectile motion, where the only external force acting on it is gravity.

Using the vertical component of the initial velocity (viy) and the time calculated above (t), we can find the height (h) at which the stream of water strikes the building.

h = viy * t + (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values we have:

h = (40.0 m/s * sin(28.0°)) * (35.0 m / (40.0 m/s * cos(28.0°))) + (1/2) * 9.8 m/s^2 * (35.0 m / (40.0 m/s * cos(28.0°)))^2

Simplifying and calculating this equation should give you the height (h) at which the stream of water strikes the building.