Answer
The vertical distance of projectile is 5.18 m below the edge
Work Step by Step
Please see the attached image first.
Let's apply the equation $V^{2}=u^{2}+2aS$ in the vertical direction to find the vertical distance.
$\downarrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V_{y}^{2}=0+2\times9.8\space m/s^{2}h$
$h=\frac{V_{y}^{2}}{19.6\space m/s^{2}}-(1)$
The x-component of the velocity of the projectile is constant. Because there is no acceleration in the horizontal direction.
So, $V_{x}=2.7\space m/s$
By using trigonometry, we can get.
$tan\theta=\frac{V_{y}}{V_{x}}$
$V_{y}=V_{x}tan75^{\circ}=2.7\space m/s\times (3.73)=10.08\space m/s$
(1)=>
$h=\frac{(10.08\space m/s)^{2}}{19.6\space m/s^{2}}=5.18\space m$
The velocity vector of the water points downward at an angle of $75^{\circ}$, below the horizontal at a vertical distance of 5.2 m below the edge.
