A) Horizontal range: 16.34 m
B) Horizontal range: 16.38 m
C) Horizontal range: 16.34 m
D) Horizontal range: 16.07 m
E) The angle that gives the maximum range is 
Explanation:
A)
The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.
The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:
(1)
where
s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)
is the initial vertical velocity, where
u = 12.0 m/s is the initial speed
is the angle of projection
So

is the acceleration due to gravity (downward)
Substituting the numbers, we get

which has two solutions:
t = -0.21 s (negative, we ignore it)
t = 1.778 s (this is the time of flight)
The horizontal motion is instead uniform, so the horizontal range is given by

where
is the horizontal velocity
t = 1.778 s is the time of flight
Solving, we find

B)
In this second case,

So the vertical velocity is

So the equation for the vertical motion becomes

Solving for t, we find that the time of flight is
t = 1.851 s
The horizontal velocity is

So, the range of the shot is

C)
In this third case,

So the vertical velocity is

So the equation for the vertical motion becomes

Solving for t, we find that the time of flight is
t = 1.925 s
The horizontal velocity is

So, the range of the shot is
m
D)
In this 4th case,

So the vertical velocity is

So the equation for the vertical motion becomes

Solving for t, we find that the time of flight is
t = 1.981 s
The horizontal velocity is

So, the range of the shot is

E)
From the previous parts, we see that the maximum range is obtained when the angle of releases is
.
The actual angle of release which corresponds to the maximum range can be obtained as follows:
The equation for the vertical motion can be rewritten as

The solutions of this quadratic equation are

This is the time of flight: so, the horizontal range is

It can be found that the maximum of this function is obtained when the angle is

Therefore in this problem, the angle which leads to the maximum range is

Learn more about projectile motion:
brainly.com/question/8751410
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