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

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.

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