An artillery shell is fired at an angle of 75.7 above the horizontal ground with an initial speed of 1720 m/s.

The acceleration of gravity is 9.8 m/s^2.
Find the total time of flight of the shell,neglecting air resistance. Answer in units of min.

t= 2vₒ•sinα/g

To find the total time of flight of the shell, we can break down the motion into horizontal and vertical components.

First, let's consider the vertical component of motion. The initial vertical velocity can be found using the given initial speed and the angle of firing.

Vertical Initial Velocity (Vy) = Initial Speed * sin(angle)

Vy = 1720 m/s * sin(75.7)

Next, we can calculate the time it takes for the shell to reach its maximum height. At the maximum height, the vertical component of velocity becomes 0.
Using the equation:

Vy = Vy0 + (-g * t)

where Vy0 is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken to reach maximum height.

Rearranging the equation:

t = (0 - Vy0) / (-g)

Since the shell starts at ground level, Vy0 is the negative of the initial vertical velocity.

t = (-Vy) / g

Now, we need to find the time it takes for the shell to fall back to the ground from its maximum height. This time is the same as the time taken to reach maximum height.

So, the total time of flight is twice the time it takes to reach the maximum height:

Total Time of Flight = 2 * t

To convert the time into minutes, we can divide the total time by 60.

Now we can calculate the values:

Vy = 1720 m/s * sin(75.7)

t = (-Vy) / g

Total Time of Flight = 2 * t / 60

Let's plug in the values and calculate: