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 $15m$ high and the water strikes the field perpendicularly, then find the value of $v_0($ in $m/s)$.

Answer 1

Answer: 10

$v_x=v_0$ (Horizontal component of velocity)
$v_y=gt\downarrow$ (Vertical component of velocity)
$\because v_x=v_y$ (At the time of strike)
$\Rightarrow gt=v_0$ ... (i)

Now,
$\Rightarrow v_0\times t=15-h$
$\frac{v_0^2}{g}=15-\frac{v_0^2}{2g}$ [From (i)]
$\Rightarrow \frac{3}{2}\frac{v_0^2}{g}=15$
$\Rightarrow v_0^2=10g$
$\Rightarrow v_0=10m/s$