Question

Water is leaving a hose at 6.8m/s . If the target is 2m away horizontally, what angle should the water leave the base.

Answer 1

hint
let the angle be \[\theta\]
now make it in to X and Y components.
acceleration is in negative Y direction which is -g.
so use equation V=U+at.
now try.......

Answer 2

let the angle be \[theta\] and the time taken to hit the target is T
X component of the velocity is \[v \cos \theta \]
then the distance between pipe and the target is \[v \cos \theta T \]
now....\[v \cos \theta T =2\]...............................(1)
Y component of the velocity is v.sin(theta) and the acceleration is -g as it is the opposite direction of velocity...
we use this equation.... \[h _{y}=u _{y}t-1/2g t ^{2}\]
here \[u _{y}=u \sin \theta \]
and after time T water will come back at it initial point..so h=0 after time T
\[0=usin \theta T-1/2g T ^{2}\]
or \[T=2usin \theta /g\].................................(2)
eliminate T from (1) and (2)...you will get the answer..
\[\theta=12.5^{0}\]

Answer 3

@souvik you shoul not tell the answers directly with out letting them to attempt solving it.