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.)
4 answers
Help will be provided if work is shown
What i have with me is
vi*sin(theta) - (g*sqrt(2*h/g))
however it is not correct.
vi*sin(theta) - (g*sqrt(2*h/g))
however it is not correct.
PHYSICS
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
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
0 answers