Answer:
665 ft
Explanation:
Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.
The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

Since the 2 sides length above make up the 200 foot monument, their total length is
0.231d + 0.07d = 200
0.301 d = 200
d = 200 / 0.301 = 665 ft
Answer:
changing the direction in which a force is exerted
To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,



Replacing at the kinetic equation to find
we have,



Applying the concepts of continuity,

We need to find A_2 then,

So the cross sectional area of the water stream at a point 0.11 m below the faucet is



Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
Answer:
The height reached is 20m, The time taken to reach 20m is 2 seconds
Explanation:
Observing the equations of motion we can see that the following equation will be most helpful for this question.

We are given initial velocity, u
We know that the stone will stop at its maximum height, so final velocity, v
Acceleration, a
And we are looking for the displacement (height reached), s
Substitute the values we are given into the equation

Rearrange for s



s = -20 (The negative is just showing direction, it can be ignored for now)
The height reached is 20m
Use a different equation to find the time taken

Substitute in the values we have

Rearrange for t



t = 2s
The time taken to reach 20m is 2 seconds