Question

A field is on the slope of a hill of inclination $45^{\circ}$ with horizontal. For irrigation purpose, a vertical pole is placed somewhere on the field, which throws a water jet with velocity $v_{0}$ horizontally in the direction shown. The plane of motion includes the line of maximum slope on the field. If the pole is $15\, m$ high and the water strikes the field perpendicularly, then find the value of $v_{0}($ in $m / s )$.

Answer 1

$v_{x}=v_{0}$ (Horizontal component of velocity)
$v_{y}=g t \downarrow$ (Vertical component of velocity)
$\because v_{x}=v_{y}$ (At the time of strike)
$\Rightarrow g t=v_{0}$ ... (i)

Now,
$\Rightarrow v_{0} \times t=15-h $
$\frac{v_{0}^{2}}{g}=15-\frac{v_{0}^{2}}{2 g}$ [From (i)]
$\Rightarrow \frac{3}{2} \frac{v_{0}^{2}}{g}=15$
$ \Rightarrow v_{0}^{2}=10 \,g $
$\Rightarrow v_{0}=10\, m / s$