An artillery shell is fired at an angle of 67.4◦ above the horizontal ground with an initial speed of 1560 m/s.

The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell, neglecting air resistance. Find its horizontal range, neglecting air resistance

Vo = 1560 m/s @ 67.4o

Range = Vo^2*sin(2A)/g
= 1560^2*sin(134.8)/9.8 = 176,205 m.

Range = Vo*cos67.4*T = 176,205
T = 176,205/1560*cos67.4 = 294 s. in
flight.

To find the total time of flight and horizontal range of the artillery shell, we can use the kinematic equations of motion. Let's break down the problem into two components: the vertical and horizontal motion.

Vertical Motion:
1. Determine the initial vertical velocity (Vy0) of the shell:
Vy0 = V0 * sin(θ)
where V0 is the initial speed of the shell (1560 m/s) and θ is the firing angle (67.4 degrees).

2. Calculate the time of flight (T) for the vertical motion:
The time it takes for the shell to reach its highest point is equal to the time it takes for the shell to fall back to the ground.
We can use the following equation:
T = (2 * Vy0) / g
where g is the acceleration due to gravity (9.8 m/s^2).

3. Calculate the vertical displacement (ΔY) of the shell:
ΔY = Vy0 * T - (1/2) * g * T^2

Horizontal Motion:
4. Determine the initial horizontal velocity (Vx0) of the shell:
Vx0 = V0 * cos(θ)

5. Calculate the horizontal range (R) of the shell:
R = Vx0 * T
where T is the time of flight obtained from the vertical motion.

Now let's plug in the values and solve the equations:

Step 1:
Vy0 = 1560 m/s * sin(67.4) = 1472.724 m/s

Step 2:
T = (2 * 1472.724 m/s) / 9.8 m/s^2 = 300.298 s (approximately)

Step 3:
ΔY = 1472.724 m/s * 300.298 s - (1/2) * 9.8 m/s^2 * (300.298 s)^2 = 221267.67 m (approximately)

Step 4:
Vx0 = 1560 m/s * cos(67.4) = 619.464 m/s

Step 5:
R = 619.464 m/s * 300.298 s = 186013.35 m (approximately)

Therefore, the total time of flight of the shell is approximately 300.298 seconds, and its horizontal range is approximately 186013.35 meters.