A firefighter a distance d from a burning building directs a stream of water from a fire hose at angle θi above the horizontal as in the figure. If the initial speed of the stream is vi, at what height h does the water strike the building? (Use theta for θi, vi for vi, g for acceleration due to gravity, and d and h as necessary.)
Vy = Visin(theta)i
Vx = Vicos(theta)i
d=Vxt
t=d/Vicos(theta)i <- substitute Vx
h=Vyt+1/2gt^2 <- since g is -9.8 it . is -g
so
h=Vyt-1/2gt^2 <- then substitute
h=Visin(theta)i x (d/Vicos(theta)i) -1/2g(d/Vicos(theta)i)^2
now simplify
h=dtan(theta)i - gd^2/2Vi^2cos^2(theta)i
To determine the height at which the water strikes the building, we need to analyze the projectile motion of the water stream. The water is ejected from the hose with an initial velocity vi at an angle θi above the horizontal.
We can break down the initial velocity into its horizontal and vertical components. The horizontal component of the initial velocity is vi * cos(θi), and the vertical component is vi * sin(θi).
Since the only force acting on the water stream is gravity, the vertical motion of the water stream can be treated as a simple projectile motion with a constant acceleration of gravity (g = 9.8 m/s^2).
Using the equations of projectile motion, we can find the time it takes for the water stream to reach the building. The time of flight (t) can be calculated using the equation:
t = 2 * (vi * sin(θi)) / g
Now, we can determine the horizontal distance covered by the water stream during this time. The horizontal distance (d) can be calculated using the equation:
d = (vi * cos(θi)) * t
Finally, we can determine the height (h) at which the water strikes the building. The height (h) can be calculated using the equation:
h = (vi * sin(θi)) * t - (1/2) * g * t^2
Now that we have the equations, you can substitute the given values of θi and vi into these equations to find the value of h.
What i have with me is
vi*sin(theta) - (g*sqrt(2*h/g))
however it is not correct.