Answer
30 m
Work Step by Step
This is a projectile motion. The horizontal velocity component remains constant throughout the motion because there isn't an acceleration in the horizontal direction.
Let's apply equation 3.5a $S=ut$ in the horizontal direction to find the displacement of the hose from the building x.
$\rightarrow S=ut$
$x=Vcos\theta\times t-(1)$
Let's apply equation 3.3b $V=u+at$ in the vertical direction to find the flight time until the water reaches its maximum height.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$0=Vsin\theta +(-g)t$
$t=\frac{Vsin\theta}{g}-(2)$
(2)=>(1),
$x=Vcos\theta\times \frac{Vsin\theta}{g}$
$x=\frac{(25\space m/s)^{2}cos35^{\circ}sin35^{\circ}}{9.8\space m/s^{2}}=30\space m$
The displacement of the hose from the building is 30 m
