To solve for the motion of the projectile launched at an angle of 45 degrees, we can use the equations of motion taking into account both air resistance and gravity.

The horizontal and vertical motions of the projectile can be described by the following equations:

Horizontal motion:
x = v0 * t + (1/2) * a * t^2

Vertical motion:
y = v0 * sin(theta) * t - (1/2) * g * t^2

Where:
x: horizontal displacement (range)
y: vertical displacement
v0: initial velocity
a: acceleration in the horizontal direction (assumed to be 0 as there is no horizontal acceleration)
theta: launch angle (45 degrees)
t: time
g: acceleration due to gravity (9.8 m/s^2)

We can solve for the time of flight by setting y = 0 (projectile hits the ground):

0 = v0 * sin(theta) * t - (1/2) * g * t^2

v0 * sin(theta) * t = (1/2) * g * t^2

v0 * sin(theta) = (1/2) * g * t

t = (2 * v0 * sin(theta)) / g

Substituting the values: v0 = 20 m/s and theta = 45 degrees, we get:

t = (2 * 20 * sin(45)) / 9.8
t ≈ 2.04 seconds

Now, we can find the range by substituting the time of flight (t) into the horizontal motion equation:

x = v0 * t + (1/2) * a * t^2

Since the horizontal acceleration (a) is 0:

x = v0 * t

x = 20 * 2.04
x ≈ 40.8 meters

Therefore, the range of the projectile is approximately 40.8 meters.

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