horizontal problem:
u = 40 cos 21 for the whole time t
d = 33
so
t = 33/(40 cos 21)
vertical problem, same t
initial vertical velocity Vi = 40 sin 21
h = Vi t - 4.9 t^2
A fireman d = 33.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 21.0° above the horizontal as shown in the figure. 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?
2 answers
Thank you so much!!!!