Question

i dont understand what i am suppose to do

Answer 1

You know the speed and the range. Start there to get the time:

Substitute into the y equation:

Since we want the water to hit the ground 2.0m away, y = 0. Substitute in for t:

<-- Typo. Should be 2tan(theta) NOT 4.Fixed below.

Use double angle identity for the sine:

2sinθcosθ=sin(2θ)

2θ=arcsin(4.905*4/6.8^2)

2θ=25.1deg

θ=12.55deg

Answer 2

Given that the speed of water is u = 6.8 m/s
The horizontal distance (range) R = 2.0m
----------------------------------------------------
Let the angle of projection is (theta) ?
Horizontal range
R = u^2 sin2θ / g
sin2θ = Rg / u^2
= (2.0m)(9.8m/s^2) / (6.8m/s)^2
=0.423
2θ = sin^-1(0.544)
2θ = 25.079
θ=12.539 degrees
Another angel is
θ' = 90 - 12.539
= 77.46 degrees