An artillery shell is fired at an angle of 66.1 degrees above the horizontal ground with an initial speed of 1650 m/s. The acceleration of gravity is 9.8 m/s2 . Find the total time of flight of the shell,neglecting air resistance.
time up equals time down (still)
initial vertical velocity is __ 1650 * sin(66.1º)
flight time is __ 2 * 1650 * sin(66.1º) / 9.8
307.8610257 minutes
To find the total time of flight of the artillery shell, we can break the projectile motion into vertical and horizontal components.
First, let's consider the vertical motion. The initial velocity in the vertical direction can be found using the initial speed and the launch angle:
Vertical velocity (V_y) = initial speed * sin launch angle
Given:
Initial speed (V_0) = 1650 m/s
Launch angle (θ) = 66.1 degrees
Therefore,
V_y = 1650 m/s * sin(66.1 degrees)
Next, we can find the time it takes for the shell to reach its highest point. At the highest point, the vertical velocity becomes zero. We can use the equation for vertical velocity:
V_y = V_0y - g * t
Since the projectile reaches its highest point, V_y becomes zero. Plugging in the known values:
0 = V_0y - g * t
Solving for time (t):
t = V_0y / g
Now, let's consider the horizontal motion. The horizontal velocity remains constant throughout the flight:
Horizontal velocity (V_x) = initial speed * cos launch angle
Given:
V_x = 1650 m/s * cos(66.1 degrees)
Since the horizontal velocity doesn't change, the time taken for the projectile to reach its highest point is equal to the time taken to fall from its highest point to the ground. Thus, the total time of flight is twice the time calculated earlier:
Total time of flight = 2 * t
Now we can substitute the values and calculate the total time of flight:
t = (1650 m/s * sin(66.1 degrees)) / 9.8 m/s^2
Total time of flight = 2 * t